Lecture Notes omplex Analysis based on omplex Variables and Applications 7th Edition Brown and hurchhill Yvette Fajardo-Lim, Ph.D. Department of Mathematics De La Salle University - Manila
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ontents THE OMPLEX NUMBER FIELD 5. Sum and Products............................... 5.2 Basic Algebraic Properties........................... 6.3 onjugate and Absolute Value of a omplex Number........... 7.4 Polar and Exponential Form...........................5 Roots of omplex Numbers.......................... 4.6 Regions in the omplex Plane......................... 20 2 ANALYTI FUNTIONS 23 2. Functions of a omplex Variable....................... 23 2.2 Limits...................................... 26 2.3 Limits Involving the Point at Infinity..................... 30 2.4 ontinuity.................................... 32 2.5 Derivatives................................... 34 2.6 Differentiation Formulas............................ 37 2.7 auchy-riemann Equations.......................... 39 2.8 Sufficient onditions for Differentiability................... 4 2.9 Polar Form of auchy-riemann Equations.................. 43 2.0 Analytic Functions............................... 44 2. Harmonic Functions.............................. 47 3 ELEMENTARY FUNTIONS 5 3. The omplex Exponential Function..................... 5 3.2 omplex Trigonometric Functions...................... 52 3.3 omplex Hyperbolic Functions........................ 54 3.4 omplex Logarithmic Functions........................ 56 3.5 omplex Inverse Trigonometric Functions.................. 59 4 INTEGRALS 6 4. Definite Integrals of w(t)............................ 6 4.2 Arcs and ontours............................... 63 4.3 ontour Integrals................................ 67 4.4 Antiderivatives................................. 7 4.5 auchy-goursat Theorem........................... 77 3
4 ONTENTS 4.6 auchy Integral Formula............................ 89 4.7 Derivatives of Analytic Functions....................... 9 4.8 Liouville s Theorem and the Fundamental Theorem of Algebra...... 92 5 SERIES 95 5. onvergence Of Sequences........................... 95 5.2 Taylor Series.................................. 00 5.3 Laurent Series.................................. 07 6 RESIDUES AND POLES 5 6. Residues..................................... 5 6.2 auchy s Residue Theorem.......................... 9 6.3 Three Types of Isolated Singular Points................... 22 6.4 Residues at Poles................................ 24 6.5 Evaluation of Improper Integrals....................... 28
hapter THE OMPLEX NUMBER FIELD. Sum and Products Definition.. A complex number, denoted by z is an ordered pair of real numbers, z = (x, y). The real number x is called the real part and the real number y is called the imaginary part of the complex number. We may write this as: x = Re z and y = Im z. The ordered pairs (x, y) are points on the complex plane, denoted by. The points on this plane with coordinates (x, 0) are points on the x-axis of and the points of the form (0, y) are points on the y-axis of. These axes are called the real axis and the imaginary axis respectively. Definition.2. Two complex numbers z = (x, y ) and z 2 = (x 2, y 2 ) are equal, that is, z = z 2 if and only if x = x 2 and y = y 2. Thus, the statement z l = z 2 means that z l and z 2 correspond to the same point in the complex plane. Definition.3. The sum z l + z 2 and the product z l z 2 of two complex numbers z l = (x l, y l ) and z 2 = (x2, y2) are defined as follows:. z + z 2 = (x + x 2, y + y 2 ); 2. z z 2 = (x x 2 y y 2, x 2 y + x y 2 ). Example.. Let z = (, 2) and z 2 = (2, 4). Find z + z 2 and z z 2. Solution. z + z 2 = ( + 2, 2 + 4) = (3, 2) z z 2 = (()(2) ( 2)(4), ((2)( 2) + ()(4)) = (0, 0) 5
6 HAPTER. THE OMPLEX NUMBER FIELD Remark.. We note that. the complex number system is an extension of the real number system; 2. we may write z = (x, y), in the form z = x + iy. This is called the rectangular form of the complex number z 3. i 2 =, where i = (0, ) Example.2. Let z = (, 2) and z 2 = (2, 4). Write z and z 2 in rectangular form and find z + z 2 and z z 2. Solution. We have z = 2i and z 2 = 2 + 4i. Hence, z + z 2 = 3 2i z z 2 = ( 2i)(2 + 4i) = 2 8i 2 = 0.2 Basic Algebraic Properties The set of complex numbers with addition (+) and multiplication ( ) given in Definition.3 satisfies the following properties.. is closed under (+) and ( ); 2. is commutative under (+) and ( ); 3. is associative under (+) and ( ); 4. for every z, z(z + z 2 ) = zz + zz 2 and (z + z 2 )z = z z + z 2 z; 5. there exists a unique zero element 0 = (0, 0), such that z + 0 = z ; 6. for every z = (x, y), there exists a unique element z = ( x, y) such that z + ( z) = 0. The element z is called the negative of z; 7. there exists a unique complex number e = (, 0), such that ze = ez = z, for every z. the element e is called the identity element of ; 8. for every z = (x, y) (0, 0), there exists a unique element z, such that zz = z z = e = (, 0). If z x = (u, v), then u = x 2 + y 2 and v = y x 2 + y2. The element z is called the multiplicative inverse of z. Remark.2. From the properties above, we can see that is a field with respect to (+) and ( ). Definition.4. Let z = x + iy and z 2 = x 2 + iy 2. Then, the difference of z and z 2 is defined as z z 2 = (x x 2 ) + i(y y 2 ). The quotient of z over z 2, z 2 0 is defined as z = z z2 = (x x 2 + y y 2 ) + i(x 2 y x y 2 ) z 2 x 2 2 +. y2 2
.3. ONJUGATE AND ABSOLUTE VALUE OF A OMPLEX NUMBER 7 Example.3. Let z = 2i and z 2 = 2 + 4i. Find z z 2 and z z 2. Solution. Exercises.. z z 2 = ( 2) + i( 2 4) = 6i z (()(2) + ( 2)(4)) + i((2)( 2) ()(4)) = z 2 2 2 + 4 2 = 6 8i = 3 4i 20 0. Verify that each of the two numbers z = ±i satisfies the equation z 2 2z +2 = 0. 2. Use the associative law for addition and the distributive law to show that z(z + z 2 + z 3 ) = zz + zz 2 + zz 3 3. By writing i = (0, ) and y = (y, 0), show that (iy) = ( i)y = i( y). 4. Solve the equation z 2 + z + = 0 for z = (x, y) by writing (x, y)(x, y) + (x, y) + (, 0) = (0, 0) and then solving a pair of simultaneous equations in x and y. 5. Reduce each of these quantities to a real number: (a) + 2i 3 4i + 2 i 5i ; (b) 5i ( i)(2 i)(3 i) ; (c) ( i) 4 6. Use the associative and commutative laws for multiplication to show that (z z 2 )(z 3 z 4 ) = (z z 3 )(z 2 z 4 ). 7. Prove that if z z 2 z 3 = 0, then at least one of the three factors is zero..3 onjugate and Absolute Value of a omplex Number Definition.5. Let z = x + iy, the complex conjugate of z, denoted by z is defined by z = x iy. The number z is represented by the point (x, y), which is the reflection in the real axis of the point (x, y) representing z (Figure.).
8 HAPTER. THE OMPLEX NUMBER FIELD Figure.: Theorem.. Let z = x + iy and z 2 = x 2 + iy 2. Then,. z + z 2 = z + z 2 ; 2. z z 2 = z z 2 ; 3. z z 2 = z z 2 ; 4. ( z z 2 ) = z z 2, z 2 0; 5. Re z = z + z 2 and Im z = z z. 2i Definition.6. The modulus or absolute value of a complex number z = x + iy, denoted by z is defined as Remark.3. z = x 2 + y 2.. The statement z < z 2 means that the point z is closer to to the origin than the point z 2 is. 2. The distance between the point z = x + iy and z 2 = x 2 + iy 2 is given by z z 2 = (x x 2 ) 2 + (y y 2 ) 2.
.3. ONJUGATE AND ABSOLUTE VALUE OF A OMPLEX NUMBER 9 Figure.2: Theorem.2. The following are some properties of the modulus of z = x + iy.. z 0; 2. zz = z 2 ; 3. z z 2 = z z 2 ; 4. z z 2 = z z 2 ; 5. Re z z and Im z z ; 6. z + z 2 z + z 2 (Triangle Inequality); 7. z + z 2 +... + z n z + z 2 +... z n ; Remark.4. The points in satisfying z z 0 = R are points lying in a circle with center z 0 and radius R.
0 HAPTER. THE OMPLEX NUMBER FIELD Example.4. Draw a sketch of the graph of z 2 + i 4. Solution. Exercises.2.. Use properties of conjugates and moduli established to show that (a) z + 3i = z 3i; (b) iz = i z; (c) (2 + i) 2 = 3 4i; (d) (2 z + 5)( 2 i) = 3 2z + 5 2. Show that (a) z z 2 z 3 = z z 2 z 3 (b) z 4 = z 4 3. Show that when z 2 and z 3 are nonzero, (a) ( z z 2 z 3 ) = z z 2 z 3 (b) z z 2 z 3 = z z 2 z 3 4. Use established properties of moduli to show that when z 3 z 4, z + z 2 z 3 + z 4 z + z 2 z 3 z 4. 5. Show that Re(2 + z + z 3 4, when z 6. Prove that
.4. POLAR AND EXPONENTIAL FORM (a) z is real if and only if z = z; (b) z is either real or pure imaginary if and only if z 2 = z 2. 7. Use mathematical induction to show that when n = 2, 3,..., (a) z + z 2 +... + z n = z + z 2 +... + z n (b) z z 2... z n = z z 2... z n 8. Let a 0, a l, a 2,..., a n, (n ) denote real numbers, and let z be any complex number. With the aid of the results in Exercise 7, show that a 0 + a l z + a 2 z 2 +... + a n z n = a 0 + a l z + a 2 z 2 +... + a n z n 9. Show that the equation z z 0 = R of a circle, centered at z 0 with radius R, can be written z 2 2 Re (z z 0 ) + z 0 2 = R 2. 0. Show that the hyperbola x 2 y 2 = bcan be written z 2 + z 2 = 2..4 Polar and Exponential Form If z = x + iy, then the polar form of z is given by z = r(cos θ + i sin θ) = rcisθ, where r = z and θ, called an argument of z, is the angle, measured in radians, that z makes with the positive real axis when z is considered as a radius vector. learly, θ = tan y. We note that if θ is an argument of z, then so are θ + 2kπ, for any k Z. x If π θ π, then θ is called the principal argument of z, and is denoted by Arg z. All other arguments are denoted by arg z. Figure.3:
2 HAPTER. THE OMPLEX NUMBER FIELD Example.5. Write the following complex numbers in their polar form:. z = + i ; 2. z = 8 Solution.. z = + i First, we find r. r = z = + i = 2 + 2 = 2 Next we find arg z. onsider tan θ = =. We take θ = π since z is in quadrant 4. Hence, + i = 2 cos π 4 + i 2 sin π 4 = 2 cos 9π 4 + i 2 sin 9π 4 = 2 cos 7π 4 i 2 sin 7π 4 2. z = 8 r = z = 8 Then arg z = π. z = 8 cos( π) + i8 sin( π) = 8 cos π i8 sin π Remark.5. The symbol e iθ or exp(iθ) is defined using Euler s formula as: e iθ = cos θ + i sin θ. Thus, we can express the complex number z = x + iy in its exponential form in the following way: z = re iθ.
.4. POLAR AND EXPONENTIAL FORM 3 Example.6. Write the following complex numbers in their exponential form:. z = + i ; 2. z = 8 Solution.. z = + i + i = 2 exp [ ] [ i π 4 ] = 2 exp i 9π 4 = [ ( 2 exp i 7π )] 4 2. z = 8 z = 8exp [i( π)] = 8exp [i(π)] We note that the equation z = re iθ is a circle with center at the origin and radius r. Thus, if r =, then the numbers e iθ lie on the unit circle with center the origin. Geometrically, we can see that: e iπ = ; e i π 2 = i; e i4π =. To consider the product and quotient of complex numbers in exponential form, we suppose z = r e iθ and z 2 = r 2 e iθ2, then and This also shows that if z = r iθ 0, then z z 2 = r r 2 e i(θ+θ2) ; z z 2 = r r 2 e i(θ θ2). z = r e iθ. Furthermore, if z = re iθ and n Z, then z n = r n e inθ.
4 HAPTER. THE OMPLEX NUMBER FIELD As a special case, let r =, then z = e iθ, thus (e iθ ) n = n e inθ = e inθ, or equivalently, we have the de Moivre s formula (cos θ + i sin θ) n = cos nθ + i sin nθ, n Z. Example.7. Evaluate ( + i) 4 and write the product in rectangular form. Solution. ( ) 4 2cis ( + i) 4 π = 4 Exercises.3. = 4cisπ = 4( + i 0) = 4. Find the principal argument Arg z when (a) z = 2. Show that i 2 2i ; (b) z = ( 3 i) 6 (a) e iθ = ; (b) e iθ = e iθ ; 3. Use de Moivre s formula to derive the following trigonometric identities (a) cos 3θ = cos 3 θ 3 cos θ sin 2 θ; (b) sin 3θ = 3 cos 2 θ sin θ sin 3 θ. 4. By writing the individual factors on the left in exponential form, performing the needed operations, and finally changing back to rectangular form, show that (a) i( i 3)( 3 + i) = 2( + i 3); (b) 5i/(2 + i) = + 2i; (c) ( + i) 7 = 8( + i); (d) ( + i 3) 0 = 2 ( + i 3) 5. Find θ, where 0 θ 2π, such that e iθ = 2..5 Roots of omplex Numbers We note that two complex numbers z = r e iθ and z 2 = r 2 e iθ2 are equal if and only if r = r 2 and θ = θ 2 + 2kπ, where k Z.
.5. ROOTS OF OMPLEX NUMBERS 5 Definition.7. Let z be a nonzero complex number and n N. A complex number z 0 is an nth root of z 0 if and only if z n = z 0. To illustrate, the complex number z 0 = i is an eighth root of z = 6. The complex number z 0 = 2 is also an eighth root of z = 6. To find the nth roots of any nonzero complex number z 0 = r 0 e iθ0, where n has one of the values n = 2, 3,..., we note that an nth root of z 0 is a nonzero number z = re iθ such that z n = z 0 or r n e inθ = z n = z 0 = r 0 e iθ0. It follows that r n = r 0 and nθ = θ 0 + 2kπ, k = 0,,..., n or equivalently, r = n r 0 and θ = θ 0 + 2kπ n = θ 0 n + 2kπ, k = 0,,..., n. n onsequently, the n roots of z 0 are the complex numbers, z = n r 0 exp [ i ( θ0 n + 2kπ )] n (k = 0, ±, ±2,...). The roots of z 0 all lie on the circle z = n r 0 about the origin and are equally spaced every 2π/n radians, starting with argument θ 0 /n. All of the distinct roots are obtained when k = 0,, 2,..., n, and no further roots arise with other values of k. We let c k (k = 0,, 2,..., n ) denote these distinct roots and write c k = n r 0 exp [ i ( θ0 n + 2kπ )] n (k = 0,, 2,..., n ). (.)
6 HAPTER. THE OMPLEX NUMBER FIELD Figure.4: We shall let z /n 0 denote the set of nth roots of z 0. If, in particular, z 0 is a positive real number r 0, the symbol r /n 0 denotes the entire set of roots; and the symbol n r 0 in equation (.) is reserved for the one positive root. When the value of θ 0 that is used in equation (.) is the principal value of arg z 0 ( π < θ 0 π), the number c 0 is referred to as the principal root. Thus when z 0 is a positive real number r 0, its principal root is n r0. In order to determine the nth roots of unity, we write in exponential form: = e i(0) and find z such that z n =. Let z = re iθ. We need to find r and θ such that z n =. We have (re iθ ) n = e i(0) r n e inθ = e i(0) Hence, r n = and nθ = 0 + 2kπ, (k = 0,, 2,..., n ) which gives us [ ( /n = n 0 exp i n + 2kπ )] (k = 0,, 2,..., n ) n Example.8. Find the three cube roots of unity. Solution. Since n = 3, we enumerate the roots for k = 0,, 2. When k = 0, [ ( z = 3 0 exp i 3 + (0)π )] 3 =
.5. ROOTS OF OMPLEX NUMBERS 7 When k =, z = 3 exp [ i ( 0 3 + 2π )] 3 = cos 2π 3 + i sin 2π 3 = 2 + i 3 2 When k = 2, z = 3 exp [ i ( 0 3 + 4π )] 3 = cos 4π 3 + i sin 4π 3 = 2 i 3 2 Example.9. Find all values of ( 8i) /3, or the three cube roots of 8i. Solution. r 0 = 8i = 8 θ 0 = π 2 From equation (.), c k = 3 8 exp [ i ( π 6 + 2kπ 3 )] (k = 0,, 2) When k = 0, c 0 = 2 exp [ i ( π 6 + (0)π 3 )] = 2 cos π 6 + i sin π 6 = 3 i
8 HAPTER. THE OMPLEX NUMBER FIELD Figure.5: Without any further calculations, from Figure.5 it is then evident that c l = 2i; and, since c 2 is symmetric to c 0, with respect to the imaginary axis, we know that c 2 = 3 i. Example.0. Find the four fourth roots of z = 8 + 8 3i. Solution. r 0 = 8 + 8 3i = 6 θ 0 = tan 8 3 8 = tan ( 3) = 2π 3 From equation (.), c k = 4 6 exp [ i ( π 6 + 2kπ )] 4 (k = 0,, 2, 3)
.5. ROOTS OF OMPLEX NUMBERS 9 When k = 0, c 0 = 2 exp [ i ( π 6 + (0)π )] 4 = 2 cos π 6 + i sin π 6 ( ) 3 = 2 2 + i 2 = 3 + i When k =, c = 2 exp [ i ( π 6 + π )] 2 = 2 cos 2π 3 + i sin 2π ( 3 = 2 ) 3 2 + i 2 = + 3i Since c 2 is symmetric to c 0, with respect to the origin, we know that c 2 = 3 i. Similarly, c 3 = 3i Exercises.4.. Find the square roots of (a) 2i; (b) 3i and express them in rectangular coordinates. 2. In each case, find all of the roots in rectangular coordinates, exhibit them as vertices of certain squares, and point out which is the principal root: (a) ( 6) /4 ; (b) ( 8 8 3i) /4 ; 3. In each case, find all of the roots in rectangular coordinates, exhibit them as vertices of certain regular polygons, and identify the principal root: (a) ( ) /3 ; (b) 8 /6
20 HAPTER. THE OMPLEX NUMBER FIELD.6 Regions in the omplex Plane An ε neighborhood z z 0 < ε of a given point z 0 consists of all points z lying inside but not on a circle centered at z 0 and with a specified positive radius ε (Figure.6). When the value of ε is understood or is immaterial in the discussion, the set z z 0 < ε is often referred to as just a neighborhood. Occasionally, it is convenient to speak of a deleted neighborhood 0 < z z 0 < ε consisting of all points z in an ε neighborhood of z 0 except for the point z 0 itself. Figure.6: A point z 0 is said to be an interior point of a set S whenever there is some neighborhood of z 0 that contains only points of S; it is called an exterior point of S when there exists a neighborhood of it containing no points of S. If z 0 is neither of these, it is a boundary point of S. A boundary point is, therefore, a point all of whose neighborhoods contain points in S and points not in S. The totality of all boundary points is called the boundary of S. The circle z =, for instance, is the boundary of each of the sets z < and z. A set is open if it contains none of its boundary points. It is left as an exercise to show that a set is open if and only if each of its points is an interior point. A set is closed if it contains all of its boundary points; and the closure of a set S is the closed set consisting of all points in S together with the boundary of S. Note that the set z < is open and z is its closure. Some sets are, of course, neither open nor closed. For a set to be not open, there must be a boundary point that is contained in the set; and if a set is not closed, there exists a boundary point not contained in the set. Observe that the punctured disk 0 < z
.6. REGIONS IN THE OMPLEX PLANE 2 is neither open nor closed. The set of all complex numbers is, on the other hand, both open and closed since it has no boundary points. An open set S is connected if each pair of points z l and z 2 in it can be joined by a polygonal line, consisting of a finite number of line segments joined end to end, that lies entirely in S. The open set z < is connected. The annulus < z < 2 is, of course, open and it is also connected (Figure.7). An open set that is connected is called a domain. Note that any neighborhood is a domain. A domain together with some, none, or all of its boundary points is referred to as a region. Figure.7: A set S is bounded if every point of S lies inside some circle z = R; otherwise, it is unbounded. Both of the sets z < and z are bounded regions, and the half plane Re z 0 is unbounded. A point z 0 is said to be an accumulation point of a set S if each deleted neighborhood of z 0 contains at least one point of S. It follows that if a set S is closed, then it contains each of its accumulation points. For if an accumulation point z 0 were not in S, it would be a boundary point of S; but this contradicts the fact that a closed set contains all of its boundary points. It is left as an exercise to show that the converse is, in fact, true. Thus, a set is closed if and only if it contains all of its accumulation points. Evidently, a point z 0 is not an accumulation point of a set S whenever there exists some deleted neighborhood of z 0 that does not contain points of S. Note that the origin is the only accumulation point of the set z n = i/n(n =, 2,...).
22 HAPTER. THE OMPLEX NUMBER FIELD Exercises.5.. Sketch the following sets and determine which are domains: (a) z 2 + i ; (b) 2z + 3 > 4; (c) Im z > ; (d) Im z = ; (e) 0 argz π/4(z 0); (f) z 4 z 2. Which sets in Exercise are neither open nor closed? 3. Which sets in Exercise are bounded? 4. In each case, sketch the closure of the set: (a) π argz π(z 0); (c) Re ( ) z 2 ; (b) Re z < z ; (d) re (z 2 ) > 0; 5. Let S be the open set consisting of all points z such that z < or z 2 <. State why S is not connected. 6. Show that a set S is open if and only if each point in S is an interior point. 7. Determine the accumulation points of each of the following sets: (a) z n, = i n (n =, 2,...); (b) z n, = i n /n(n =, 2,...); (c) 0 argz π/2(z 0); (d) z n, = ( ) n ( + i) n n n (n =, 2,...) 8. Prove that if a set contains each of its accumulation points, then it must be a closed set. 9. Show that any point z 0 of a domain is an accumulation point of that domain. 0. Prove that a finite set of points z l, z 2,..., z n cannot have any accumulation points.
hapter 2 ANALYTI FUNTIONS We now consider functions of a complex variable and develop a theory of differentiation for them. The main goal of the chapter is to introduce analytic functions, which play a central role in complex analysis. 2. Functions of a omplex Variable Let S be a set of complex numbers. A function f defined on S is a rule that assigns to each z in S a complex number w. The number w is called the value of f at z and is denoted by f(z); that is, w = f(z). The set S is called the domain of definition of f. We let z = x + iy and f(z) = w = u + iv, where u = Re f and v = Im f. We can see that u and v are real valued functions of the real variables x and y. Example 2.. Find the real and imaginary parts of f(z) = z 2. Solution. f(z) = z 2 = (x + iy) 2 = (x 2 y 2 ) + 2xyi. Thus, u(x, y) = x 2 y 2 and v(x, y) = 2xy. Example 2.2. Find u and v for f(z) = z. Solution. f(z) = x + iy x iy x iy = x iy x 2 + y 2 x Thus, u(x, y) = x 2 and v(x, y) = y + y2 x 2 + y 2. Properties of a real-valued function of a real variable are often exhibited by the graph of the function. But when w = f(z), where z and w are complex, no such convenient graphical representation of the function f is available because each of the numbers z and w is located in a plane rather than on a line. One can, however, display some information about the function by indicating pairs of corresponding points z = (x, y) andw = (u, v). To do this, it is generally simpler to draw the z and w planes separately. 23
24 HAPTER 2. ANALYTI FUNTIONS When a function f is thought of in this way, it is often referred to as a mapping, or transformation. The image of a point z in the domain of definition S is the point w = f(z), and the set of images of all points in a set T that is contained in S is called the image of T. The image of the entire domain of definition S is called the range of f. The inverse image of a point w is the set of all points z in the domain of definition of f that have w as their image. The inverse image of a point may contain just one point, many points, or none at all. The last case occurs, of course, when w is not in the range of f. Example 2.3. From Example 2., sketch the mapping of f(z) from the z plane to the w plane. Solution. The mapping w = z 2 can be thought of as the transformation u(x, y) = x 2 y 2, v(x, y) = 2xy from the xy plane to the uv plane. Each branch of a hyperbola x 2 y 2 = c, c > 0 is mapped in a one to one manner onto the vertical line u = c l. We start by noting from u(x, y) = x 2 y 2 that u = c l when (x, y) is a point lying on either branch. When, in particular, it lies on the right-hand branch, v(x, y) = 2xy tells us that v = 2y y 2 + c. Thus the image of the right-hand branch can be expressed parametrically as u = c, v = 2y y 2 + c ( < y < ); and it is evident that the image of a point (x, y) on that branch moves upward along the entire line as (x, y) traces out the branch in the upward direction (Figure 2.). Likewise, since the pair of equations u = c, v = 2y y 2 + c ( < y < ); furnishes a parametric representation for the image of the left-hand branch of the hyperbola, the image of a point going downward along the entire left-hand branch is seen to move up the entire line u = c. Figure 2.: w = z 2
2.. FUNTIONS OF A OMPLEX VARIABLE 25 On the other hand, each branch of a hyperbola 2xy = c 2, c 2 > 0 is transformed into the line v = c 2, as indicated in Figure 2.. To verify this, we note from v(x, y) = 2xy that v = c 2 when (x, y) is a point on either branch. Suppose that it lies on the branch lying in the first quadrant. Then, since y = c 2 /(2x), u(x, y) = x 2 y 2 reveals that the branch s image has parametric representation u = x 2 c2 2 4x 2, v = c 2 (0 < x < ). Since u depends continuously on x, then, it is clear that as (x, y) travels down the entire upper branch of hyperbola 2xy = c 2, c 2 > 0, its image moves to the right along the entire horizontal line v = c 2. Inasmuch as the image of the lower branch has parametric representation u = c2 2 4y 2 y2, v = c 2 ( < y < 0), the image of a point moving upward along the entire lower branch also travels to the right along the entire line v = c 2 (see Figure 2.). The next example illustrates how polar coordinates can be useful in analyzing certain mappings. Example 2.4. Sketch the graph of f(z) = z 2 using polar coordinates. Solution. f(z) = z 2 = r 2 (e iθ ) 2 = r 2 e 2iθ Hence, U = {z = re iθ : r 0, 0 θ π/2} and V = {w = ρe iθ : ρ 0, 0 θ π} Figure 2.2: w = z 2 Observe that points z = r 0 e iθ on a circle r = r 0 are transformed into points w = r 2 e 2iθ on the circle ρ = r 2 0. As a point on the first circle moves counterclockwise from the positive real axis to the positive imaginary axis, its image on the second circle moves counterclockwise from the positive real axis to the negative real axis (see Figure 2.2).
26 HAPTER 2. ANALYTI FUNTIONS So, as all possible positive values of r 0 are chosen, the corresponding arcs in the z and w planes fill out the first quadrant and the upper half plane, respectively. The transformation w = z 2 is, then, a one to one mapping of the first quadrant r 0, 0 θ π/2 in the z plane onto the upper half ρ 0, 0 θ π of the w plane, as indicated in Figure 2.2. The point z = 0 is, of course, mapped onto the point w = 0. 2.2 Limits Recall the limit definition of functions of a real variable. We say that if ɛ > 0, δ > 0, such that lim f(x) = L, x x 0 f(x) L < ɛ whenver 0 < x x 0 < δ. We extend this notion to functions of a complex variable. Definition 2.. Let f be a function of a complex variable z, defined on some set containing the complex number z 0. We say that the limit of f(z) as z approaches z 0 is w 0, written as lim z z 0 f(z) = w 0, if for every ε > 0, there exists a δ > 0, such that f(z) w 0 < ε whenever 0 < z z 0 < δ. Geometrically, this definition says that, for each ε neighborhood w w 0 < ɛ of w 0, there is a deleted δ neighborhood 0 < z z 0 < δ of z 0 such that every point z in it has an image w lying in the ε neighborhood (Figure 2.3). Note that even though all points in the deleted neighborhood 0 < z z 0 < δ are to be considered, their images need not fill up the entire neighborhood w w 0 < ɛ. If f has the constant value w 0, for instance, the image of z is always the center of that neighborhood. Note, too, that once a δ has been found, it can be replaced by any smaller positive number, such as δ/2. Figure 2.3:
2.2. LIMITS 27 Example 2.5. Use the definition of limits to show that lim (3z i) = 3 + 2i. z +i Solution. hoose ε > 0. We want to find δ > 0 such that whenever 0 < z (+i) < δ, then f(z) (3 + 2i) < ε. f(z) (3 + 2i) < ε 3z i (3 + 2i) < ε 3z 3 i 2i < ε 3z 3 3i < ε 3z 3( + i) < ε 3(z ( + i)) < ε z ( + i) < ε 3 We take δ < ε 3. z ( + i) < δ z ( + i) < ε 3 3 z ( + i) < ε 3(z ( + i)) < ε 3z 3( + i) < ε f(z) (3 + 2i) < ε Theorem 2.. The limit of a function f(z), if it exists, is unique. Proof. Let lim z z0 f(z) = w 0 and lim z z0 f(z) = w. Then, for any positive number ε, there are positive numbers δ 0 and δ, such that and f(z) w 0 < ε whenever 0 < z z 0 < δ 0 f(z) w 0 < ε whenever 0 < z z 0 < δ. So if 0 < z z 0 < δ, where δ denotes the smaller of the two numbers δ 0 and δ, we find that w 0 w = w 0 f(z) + f(z) w = ((f(z) w 0 ) + (f(z) w ) ((f(z) w 0 ) + (f(z) w ) < ε + ε = 2ε But w 0 w is a nonnegative constant, and ε can be chosen arbitrarily small. Hence, w w 0 = 0.
28 HAPTER 2. ANALYTI FUNTIONS z Example 2.6. Show that lim z 0 z Solution. does not exist. Figure 2.4: Along the real axis, z = x + 0i and f(z) = x + 0i =. Hence, lim f(z) = lim x 0i =. z 0 x 0 Along the imaginary axis, y 0, z = 0 + iy and f(z) = 0 + iy =. Hence, lim lim y 0 0 iy f(z) = z 0 ( ) =. Since the limit is supposed to be unique if it exists, we conclude that z lim z 0 does not exist. z Theorem 2.2. Let z = x + iy and z 0 = x 0 + iy 0. Let f(z) = w = u + iv and f(z 0 ) = w 0 = u 0 + iv 0. Then, lim f(z) = w 0, z z 0 if and only if Proof. ( ) Suppose lim u(x, y) = u 0 and (x,y) (x 0,y 0) lim v(x, y) = v 0. (x,y) (x 0,y 0) lim u(x, y) = u 0 and (x,y) (x 0,y 0) positive number ε, there exist positive numbers δ and δ 2 such that lim v(x, y) = v 0. Hence, for each (x,y) (x 0,y 0) and u u 0 < ɛ 2 whenever 0 < (x x 0 ) 2 + (y y 0 ) 2 < δ v v 0 < ɛ 2 whenever 0 < (x x 0 ) 2 + (y y 0 ) 2 < δ 2.
2.2. LIMITS 29 Let δ denote the smaller of the two numbers δ and δ 2. Since (u + iv) (u 0 + iv 0 ) = (u u 0 ) + i(v v 0 ) (u u 0 ) + (v v 0 ) and (x x0 ) 2 + (y y 0 ) 2 = (x x 0 ) + i(y y 0 ) = (x + iy) (x 0 + iy 0 ), it follows that (u + iv) (u 0 + iv 0 ) < ε 2 + ε 2 = ε whenever Therefore, lim z z 0 f(z) = w 0. 0 < (x + iy) (x 0 + iy 0 ) < δ. ( ) Suppose lim z z 0 f(z) = w 0. Hence, for each positive number ε, there is a positive number δ such that (u + iv) (u 0 + iv 0 ) < ε whenever But and 0 < (x + iy) (x 0 + iy 0 ) < δ. u u 0 (u u 0 ) + i(v v 0 ) = (u + iv) (u 0 + iv 0 ), v v 0 (u u 0 ) + i(v v 0 ) = (u + iv) (u 0 + iv 0 ), (x + iy) (x 0 + iy 0 ) = (x x 0 ) + i(y y 0 ) = (x x 0 ) 2 + (y y 0 ) 2. It follows that whenever u u 0 < ε and v v 0 < ε 0 < (x x 0 ) 2 + (y y 0 ) 2 < δ. Theorem 2.3. Suppose that Then lim z z 0 f(z) = w 0 and. lim z z 0 (f(z) + F (z)) = w 0 + W 0 ; 2. lim z z 0 (f(z)f (z)) = w 0 W 0 ; f(z) 3. lim z z 0 F (z) = w 0, if W 0 0. W 0 lim z z 0 F (z) = W 0.
30 HAPTER 2. ANALYTI FUNTIONS Example 2.7. Evaluate the following limits.. lim z 3i z 2 + 9 z 3i 2. lim z i z 2 + z 4 i Solution.. lim z 3i z 2 + 9 z 3i z 2 + 9 z 3i (z + 3i)(z 3i) = z 3i = z + 3i Hence, lim z 3i z 2 + 9 z 3i = 6i 2. lim z i z 2 + z 4 i = 0 Theorem 2.4. Let P (z) = a 0 + a z + a 2 z 2 +... + a n z n. Then, lim P (z) = P (z 0 ). z z 0 2.3 Limits Involving the Point at Infinity It is sometimes convenient to include with the complex plane the point at infinity, denoted by, and to use limits involving it. The complex plane together with this point is called the extended complex plane, To visualize the point at infinity, one can think of the complex plane as passing through the equator of a unit sphere centered at the point z = 0 (Figure 2.5). To each point z in the plane there corresponds exactly one point P on the surface of the sphere. The point P is determined by the intersection of the line through the point z and the north pole N of the sphere with that surface. In like manner, to each point P on the surface of the sphere, other than the north pole N, there corresponds exactly one point z in the plane. By letting the point N of the sphere correspond to the point at infinity, we obtain a one to one correspondence between the points of the sphere and the points of the extended complex plane. The sphere is known as the Riemann sphere, and the correspondence is called a stereographic projection.
2.3. LIMITS INVOLVING THE POINT AT INFINITY 3 Figure 2.5: Observe that the exterior of the unit circle centered at the origin in the complex plane corresponds to the upper hemisphere with the equator and the point N deleted. Moreover, for each small positive number ε those points in the complex plane exterior to the circle z = /ε correspond to points on the sphere close to N. We thus call the set z > /ε an ε neighborhood, or neighborhood of. In referring to a point z, we mean a point in the finite plane. Hereafter, when the point at infinity is to be considered, it will be specifically mentioned. Theorem 2.5. If z 0 and w 0 are points in the z and w planes respectively, then and lim f(z) = if and only if z z 0 lim f(z) = w 0 if and only if lim f z z 0 lim z z 0 f(z) = 0 ( ) = w 0. z Moreover, lim f(z) = if and only if lim z z 0 f(/z) = 0. Example 2.8. Evaluate the following limits:. lim z Solution. iz + 3 ; 2. lim z + z iz + 3. lim z z + = since lim z 2z + i 2z 3 ; 3. lim z + z z 2 + ; z + iz + 3 = 0 ; 2z + i (2/z) + i 2. lim = 2 since lim z z + z 0 (/z) + = lim 2 + iz z 0 + z = 2; 2z 3 (/z 2 ) + 3. lim z z 2 = since lim + z 0 (2/z 3 ) = lim z + z 3 z 0 2 z 3 = 0;
32 HAPTER 2. ANALYTI FUNTIONS 2.4 ontinuity Definition 2.2. A function f is continuous at a point z 0, if the following conditions are satisfied:. lim z z 0 f(z) exists; 2. f(z 0 ) exists; 3. lim z z 0 f(z) = f(z 0 ) We can see that statement (3) above contains () and (2). Thus, if f(z) = u + iv, then f is continuous at z 0 if and only if lim f(z) = f(z 0 ) z z 0 = u(x 0, y 0 ) + v(x 0, y 0 ) = lim u(x, y) + lim (x,y) (x 0,y 0) (x,y) (x 0,y 0) v(x, y) Therefore, f is continuous at z 0 if and only if u and v are both continuous at (x 0, y 0 ). Definition 2.3. A function f is continuous in a region R if it is continuous at every point in R. Theorem 2.6. A composition of continuous functions is continuous. Proof. Let w = f(z) be a function that is defined for all z in a neighborhood z z 0 < δ of a point z 0, and we let W = g(w) be a function whose domain of definition contains the image of that neighborhood under f. The composition W = g[f(z)] is, then, defined for all z in the neighborhood z z 0 < δ. Suppose now that f is continuous at z 0 and that g is continuous at the point f(z 0 ) in the w plane. In view of the continuity of g at f(z 0 ), there is, for each positive number ε, a positive number γ such that g[f(z)] g[f(z 0 )] < ε whenever 0 < f(z) f(z 0 ) < γ. (See Figure 2.6.) But the continuity of f at z 0 ensures that the neighborhood z z 0 < δ can be made small enough that the second of these inequalities holds. The continuity of the composition g[f(z)] is, therefore, established.
2.4. ONTINUITY 33 Figure 2.6: Theorem 2.7. If a function f(z) is continuous and nonzero at a point z 0, then f(z) 0, throughout some neighborhood of z 0. Proof. Assume that f(z) is continuous and nonzero at a point z 0. Suppose f(z) = 0. Assign the positive value f(z 0 ) /2 to the number ε in f(z) f(z 0 ) < ε whenever 0 < z z 0 < δ. This tells us that there is a positive number δ such that f(z) f(z 0 ) < f(z 0) 2 whenever 0 < z z 0 < δ. So if there is a point z in the neighborhood z z 0 < δ at which f(z) = 0, we have the contradiction f(z 0 ) < f(z 0) ; 2 and the theorem is proved. Exercises 2... Use Definition 2. to prove that (a) lim z z 0 Re z = Re z 0 ; (b) lim z z 0 z = z 0 ; (c) z 2 lim z z 0 z = 0. 2. Let a, b, and c denote complex constants. Then use Definition 2. to show that (a) lim (az + b) = az 0 + b; z z 0
34 HAPTER 2. ANALYTI FUNTIONS (b) (c) lim (z 2 + c) = z0 2 + c; z z 0 lim [x + i(2x + y)] = + i, z i (z = x + iy).. 3. Let n be a positive integer and let P (z) and Q(z) be polynomials, where Q(z 0 ) 0. Find (a) lim z z 0 z n (z iz 3 P (z) 0 0); (b) lim ; (c) lim z i z + i z z 0 Q(z). 4. Write z = z z 0, and show that lim f(z) = w 0 if and only if z z 0 lim f(z 0 + z) = w 0. z 0 5. Show that if lim z z 0 f(z) = 0 and that if there exists a positive number M such that g(z) M for all z in some neighborhood of z 0 then lim z z 0 f(z)g(z) = 0. 6. Evaluate the following: (a) lim z 4z 2 z 2 + (z ) 2 (b) lim (c) lim z (z ) 3 z z 2.5 Derivatives Definition 2.4. Let f be a function whose domain of definition contains a neighborhood of a point z 0. The derivative of f at z 0, denoted by f (z 0 ) is defined as f (z 0 ) = lim z z 0 f(z) f(z 0 ) z z 0, (2.) provided this limit exists. The function f is said to be differentiable at z 0 when f (z 0 ) exists. If we let z = z z 0, then Equation 2., can be written as f (z 0 ) = lim z 0 f(z 0 + z) f(z 0 ). (2.2) z Note that, because f is defined throughout a neighborhood of z 0, the number f(z 0 + z) is always defined for z sufficiently small (Figure 2.7).
2.5. DERIVATIVES 35 Figure 2.7: Letting w = f(z + z) f(z), we may write f (z) as dw and Equation 2.5 becomes dz dw dz = lim w z 0 z (2.3) Example 2.9. Let f(z) = z 2, find f (z) for any z. Solution. f w (z) = lim z 0 z = lim (z + z) 2 z 2 = lim (2z + z) = 2z. z 0 z z 0 Example 2.0. Let f(z) = z 2, can we find f (z)? Solution. w z = z + z 2 z 2 z (z + z)(z + z) z z = z (z + z)( z + z) z z = z z z + z z + z( z) + ( z)( z) z z = z = z z + z + z z
36 HAPTER 2. ANALYTI FUNTIONS Figure 2.8: If the limit of w exists, it may be found by letting the point z = ( x, y) approach z the origin in the z plane in any manner. In particular, when z approaches the origin horizontally through the points ( x, 0) on the real axis (Figure 2.8), lim z 0 [ z z ] + z + z z [ ] x i0 = lim z + z + x i0 z 0 x + i0 = lim [z + z + x] x 0 = z + z When z approaches the origin vertically through the points (0, y) on the imaginary axis, [ lim z z ] [ + z + z = lim z 0 i y ] + z + 0 i y z 0 z z 0 0 + i y Since limits are unique, it follows that or z = 0, if dw/dz is to exist. = lim [ z + z i y] y 0 = z z z + z = z z Example 2.0 shows that a function can be differentiable at a certain point but nowhere else in any neighborhood of that point. Since the real and imaginary parts of f(z) = z 2 are u(x, y) = x 2 + y 2 and v(x, y) = 0,
2.6. DIFFERENTIATION FORMULAS 37 respectively, it also shows that the real and imaginary components of a function of a complex variable can have continuous partial derivatives of all orders at a point and yet the function may not be differentiable there. The function f(z) = z 2 is continuous at each point in the plane since its components are continuous at each point. So the continuity of a function at a point does not imply the existence of a derivative there. It is, however, true that the existence of the derivative of a function at a point implies the continuity of the function at that point. Remark 2.. To show that a function f is not differentiable at z 0 evaluate the limit f (z 0 ) = lim z 0 f(z 0 + z) f(z 0 ), z from two different directions such that the resulting limits are different. Example 2.. Let f(z) = z, show that f (z) does not exist, for any z. Solution. w z z + z z = z z + z z = z = z z When z approaches the origin horizontally through the points ( x, 0) on the real axis, z lim z 0 z x i0 = lim x 0 x + i0 =. However, when z approaches the origin vertically through the points (0, y) on the imaginary axis, z lim z 0 z 0 i y = lim y 0 0 + i y = Since limits are unique, it follows that dw/dz does not exist. 2.6 Differentiation Formulas The following are some differentiation rules of a function f of a complex variable.. d c = 0, where c is a complex constant; dz
38 HAPTER 2. ANALYTI FUNTIONS 2. 3. d dz z = ; d dz [cf(z)] = cf (z), where c is a complex constant; 4. Let n Z +, then d dz zn = nz n. This formula is still valid even when n Z, provided z 0. 5. Suppose the derivatives of f(z) and g(z) exists at a point z, then where g(z) 0 d dz [f(z) + g(z)] = f (z) + g (z); d dz [f(z)g(z)] = f(z)g (z) + f (z)g(z); [ ] d f(z) = g(z)f (z) f(z)g (z) dz g(z) [g(z)] 2, 6. (hain Rule) Suppose f (z 0 ) exists and g (f(z 0 )) exists, then if F (z) = g(f(z)), then F (z 0 ) = g (f(z 0 ))f (z 0 ). If we write w = f(z) and W = g(w), then W = F (z), then dw dz = dw dw Example 2.2. Let f(z) = + z z, find f (z). Solution. dw dz. f (z) = = ( z) + ( + z) ( z) 2 2 ( z) 2 Example 2.3. Let f(z) = (2z 2 + i) 5, find f (z). Solution. f (z) = 5(2z 2 + i) 4 (4z) = 20z(2z 2 + i) 4
2.7. AUHY-RIEMANN EQUATIONS 39 Exercises 2.2.. Use the appropriate differentiation rule to find f (z) when (a) f(z) = 3z 2 2z + 4; (b) f(z) = ( 4z 2 ) 3 ; (c) f(z) = z, (z /2); 2z + (d) f(z) = ( + z2 ) 4 z 2, (z 0) 2. Apply the definition of derivatives to give a direct proof that f (z) = z 2 when f(z) =, (z 0). z 3. Derive the derivative rule of sum of two functions of a complex variable. 4. Derive the formula for the derivative of the function f(z) = z n, when n is a positive integer by using (a) mathematical induction and the derivative rule for the product of two functions; (b) the definition of the derivative and the binomial formula. 5. Use the method in Example 2.0, to show that f (z) does not exist at any point z when (a) f(z) = Re z; (b) f(z) = Im z 2.7 auchy-riemann Equations In this section, we obtain a pair of equations that the first-order partial derivatives of the component functions u and v of a function f(z) = u(x, y) + iv(x, y) must satisfy at a point z 0 = (x 0, y 0 ) when the derivative of f exists there. We also show how to express f (z 0 ) in terms of those partial derivatives. We start by writing z 0 = x 0 + iy 0, z = x + i y, and w = f(z 0 + z) f(z 0 ) = [u(x 0 + x, y 0 + y) u(x 0, y 0 )] + i[v(x 0 + x, y 0 + y) v(x 0, y 0 )]. Assuming that the derivative f (z 0 ) = lim z 0 w z
40 HAPTER 2. ANALYTI FUNTIONS exists, then and f (z 0 ) = w lim Re ( x, y) (0,0) z + lim ( x, y) (0,0) Im w z. (2.4) Now, we let ( x, y) approach (0, 0) horizontally through the points ( x, 0), w lim Re ( x, y) (0,0) z = lim u(x 0 + x, y 0 ) u(x 0, y 0 ) = u x (x 0, y 0 ) x 0 x w lim Im ( x, y) (0,0) z = lim v(x 0 + x, y 0 ) v(x 0, y 0 ) = v x (x 0, y 0 ) x 0 x where u x (x 0, y 0 ) and v x (x 0, y 0 ) denote the first-order partial derivatives with respect to x of the functions u and v, respectively, at (x 0, y 0 ). Substitution of these limits into Equation 2.4 tells us that and f (z 0 ) = u x (x 0, y 0 ) + iv x (x 0, y 0 ). We let ( x, y) approach (0, 0) vertically through the points (0, y). w lim Re ( x, y) (0,0) z = lim v(x 0, y 0 + y) v(x 0, y 0 ) = v y (x 0, y 0 ) y 0 y w lim Im ( x, y) (0,0) z = lim u(x 0, y 0 + y) u(x 0, y 0 ) = u y (x 0, y 0 ) y 0 y It follows from Equation 2.4 that f (z 0 ) = v y (x 0, y 0 ) iu y (x 0, y 0 ), where the partial derivatives of u and v are, this time, with respect to y. Equating the real and imaginary parts on the right-hand sides of the above equations, we see that the existence of f (z 0 ) requires that u x (x 0, y 0 ) = v y (x 0, y 0 ) and v x (x 0, y 0 ) = u y (x 0, y 0 ). (2.5) Equations 2.5 are the auchy-riemann equations, so named in honor of the French mathematician A. L. auchy (789-857), who discovered and used them, and in honor of the German mathematician G. F. B. Riemann (826-866), who made them fundamental in his development of the theory of functions of a complex variable. Theorem 2.8. Suppose that f(z) = u(x, y) + iv(x, y) and that f (z) exists at a point z 0 = x 0 + iy 0. Then, the first order partial derivatives of u and v must exists at (x 0, y 0 ), and must satisfy the auchy-riemann equations. u x = v y and u y = v x (2.6)
2.8. SUFFIIENT ONDITIONS FOR DIFFERENTIABILITY 4 there. Also, f (z 0 ) can be written as f (z 0 ) = u x + iv x, where the partial derivatives are to be evaluated at (x 0, y 0 ). Example 2.4. Let f(z) = z 2. We have shown that f is differentiable everywhere and that f (z) = 2z. Show that the auchy-riemann equations are satisfied. Solution. Since z 2 = x 2 y 2 + i2xy, we have Thus, Moreover, u(x, y) = x 2 y 2 and v(x, y) = 2xy. u x = 2x = v y, u y = 2y = v x. f (z) = 2x + i2y = 2(x + iy) = 2z. Remark 2.2. The auchy-riemann equations are necessary conditions for differentiability of f at z 0. Thus, these equations are used to locate points in which f do not have a derivative. Example 2.5. Let f(z) = z 2. Use the auchy-riemann equations to show that f (z) is nonexistent for any nonzero z. Solution. Since f(z) = z 2, we have u(x, y) = x 2 + y 2 and v(x, y) = 0. If the auchy-riemann equations are to hold at a point (x, y), it follows that 2x = 0 and 2y = 0, or that x = y = 0. onsequently, f (z) does not exist at any nonzero point, as we already know from Example 2.0. 2.8 Sufficient onditions for Differentiability Satisfaction of the auchy-riemann equations at a point z 0 = (x 0, y 0 ) is not sufficient to ensure the existence of the derivative of a function f(z) at that point. But, with certain continuity conditions, we have the following useful theorem. Theorem 2.9. (auchy-riemann Equations) Let the function f(z) = u(x, y) + iv(x, y), be defined throughout some ε neighborhood of a point z 0 = x 0 + iy 0, and suppose that the first order partial derivatives of the functions u and v exists everywhere in that neighborhood. If those partial derivatives are continuous at (x 0, y 0 ) and satisfy the auchy- Riemann equations at (x 0, y 0 ), then f (z 0 ) exists. u x = v y and u y = v x
42 HAPTER 2. ANALYTI FUNTIONS Proof. We write z = x + i y, where 0 < z < ε, and Thus, where and w = f(z 0 + z) f(z 0 ). w = u + i v, u = u(x 0 + x, y 0 + y) u(x 0, y 0 ) v = v(x 0 + x, y 0 + y) v(x 0, y 0 ). The assumption that the first-order partial derivatives of u and v are continuous at the point (x 0, y 0 ) enables us to write and u = u x (x 0, y 0 ) x + u y (x 0, y 0 ) y + ε ( x)2 + ( y) 2 v = v x (x 0, y 0 ) x + v y (x 0, y 0 ) y + ε 2 ( x)2 + ( y) 2, where ε and ε 2 approaches 0 as ( x, y) approaches (0, 0) in the z plane. By substitution, ] w = u x (x 0, y 0 ) x+u y (x 0, y 0 ) y+ε ( x)2 + ( y) 2 +i [v x (x 0, y 0 ) x + v y (x 0, y 0 ) y + ε 2 ( x)2 + ( y) 2. (2.7) Assuming that the auchy-riemann equations are satisfied at (x 0, y 0 ), we can replace u y (x 0, y 0 ) by v x (x 0, y 0 )) and v y (x 0, y 0 ) by u x (x 0, y 0 ) in Equation 2.7 and then divide through by z to get w z = u x(x 0, y 0 ) + iv x (x 0, y 0 ) + (ε + iε 2 ) But ( x) 2 + ( y) 2 = z,and so ( x)2 + ( y) 2 z = ( x)2 + ( y) 2. (2.8) z Also, ε + iε 2 approaches 0 as ( x, y) approaches (0, 0). So the last term on the right in Equation 2.8 approaches 0 as the variable z = x + i y approaches to 0. This means that the limit of the left-hand side of Equation 2.8 exists and that f (z 0 ) = u x + iv x where the right-hand side is to be evaluated at (x 0, y 0 ). Example 2.6. Show that the derivative of f where f(z) = e z exists everywhere in. Solution. We have f(z) = e z = e x + e iy = e x cos y + ie x sin y. Then u(x, y) = e x cos y and v(x, y) = e x sin y. Since u x = v y, and u y = v x everywhere and since these derivatives are everywhere continuous, the conditions in the theorem are satisfied at all points in the complex plane. Thus f (z) exists everywhere, and f (z) = u x + iv x = e x cos y + ie x sin y.
2.9. POLAR FORM OF AUHY-RIEMANN EQUATIONS 43 2.9 Polar Form of auchy-riemann Equations Recall that x = r cos θ and y = r sin θ. Suppose z 0, then if z = re iθ, we can express f(z) as f(z) = u(r, θ) + iv(r, θ). Suppose the first-order partial derivatives of u and v with respect to x and y exist everywhere in an ɛ-neighborhood of z 0 and are continuous at that point. Then, the firstorder partial derivatives of u and v with respect to r and θ also have these properties. Moreover, u r = u x x r + u y y r, u θ = u x x θ + u y y θ ; or equivalently, u r = u x cos θ + u y sin θ, u θ = u x r sin θ + u y r cos θ. Similarly, v r = v x cos θ + v y sin θ, v θ = v x r sin θ + v y r cos θ. Theorem 2.0. Let the function f(z) = u(r, θ) + iv(r, θ) be defined throughout some ε neighborhood of a nonzero point z 0 = r 0 e iθ0, and suppose the first-order partial derivatives of u and v with respect to r and θ exist everywhere in that neighborhood. If those first-order partial derivatives are continuous at (r 0, θ 0 ) and satisfy ru r = v θ ; u θ = rv r, at the point (r 0, θ 0 ), then f (z 0 ) exists. The derivative f (z 0 ) here can be written f (z 0 ) = e iθ (u r + iv r ) where the right-hand side is to be evaluated at (r 0, θ 0 ), Example 2.7. Let f(z) = r 2 (cos 2θ + i sin 2θ). Determine where f (z) exists. Solution. Since f(z) = r 2 (cos 2θ + i sin 2θ), we have u r = 2r cos 2θ and v r = 2r sin 2θ. Hence, f (z) = e iθ (u r + iv r ) = (cos θ i sin θ)(2r cos 2θ + i2r sin 2θ) = 2r(cos θ i sin θ)(cos 2θ + i sin 2θ) = 2r(cos θ cos 2θ + sin θ sin 2θ) + i(sin 2θ cos θ cos 2θ sin θ) = 2r[cos(2θ θ) + i sin(2θ θ)] = 2r(cos θ + i sin θ)
44 HAPTER 2. ANALYTI FUNTIONS Exercises 2.3.. Use Theorem 2.8 to show that f (z) does not exist at any point. (a) f(z) = z; (b) f(z) = z z; (c) f(z) = 2x + ixy 2 ; (d) f(z) = e x e iy 2. Use Theorem 2.9 to show that f (z) and its derivative f (z) exists everywhere, and find f (z) when (a) f(z) = iz + 2; (b) f(z) = e z ; (c) f(z) = z 3 ; (d) f(z) = cos x cosh y i sin x sinh y 3. Use Theorem 2.8 and Theorem 2.9, to determine where f (z) exists and find its value when (a) f(z) = z ; (b) f(z) = x 2 + iy 2 ; (c) f(z) = zim z 4. Use Theorem 2.0 to show that each of these functions is differentiable in the indicated domain of definition, and then find f (z). (a) f(z) = /z 4, (z 0); (b) f(z) = re iθ/2, (r > 0, α < θ < α + 2π); (c) f(z) = e θ cos(ln r) + ie θ sin(ln r), (r > 0, 0 < θ < 2π) 2.0 Analytic Functions Definition 2.5. A function f of a complex variable z is said to be analytic in an open set if it has a derivative at each point in that set. If we should speak of a function f that is analytic in a set S which is not open, it is to be understood that f is analytic in an open set containing S. In particular, f is analytic at a point z 0 if it is analytic throughout some neighborhood of z 0. We note, for instance, that the function f(z) = /z is analytic at each nonzero point in the finite plane. But the function f(z) = z 2 is not analytic at any point since its derivative exists only at z = 0 and not throughout any neighborhood. Remark 2.3. The terms regular and holomorphic is synonymous to analytic. Definition 2.6. Let f be a function of a complex variable z and z 0.. If f is analytic in, then f is called an entire function.
2.0. ANALYTI FUNTIONS 45 2. If f is not analytic at z 0, but is analytic in the open neighborhood B(z 0, ɛ)\{z 0 }, then z 0 is called a singular point of f or a singularity of f. 3. A function is analytic in a domain D, if it is analytic at every point in D. The point z = 0 is evidently a singular point of the function f(z) = /z. The function f(z) = z 2, on the other hand, has no singular points since it is nowhere analytic. A necessary, but by no means sufficient, condition for a function f to be analytic in a domain D is clearly the continuity of f throughout D. Satisfaction of the auchy- Riemann equations is also necessary, but not sufficient. The derivatives of the sum and product of two functions exist wherever the functions themselves have derivatives. Thus, if two functions are analytic in a domain D, their sum and their product are both analytic in D. Similarly, their quotient is analytic in D provided the function in the denominator does not vanish at any point in D. In particular, the quotient P (z)/q(z) of two polynomials is analytic in any domain throughout which Q(z) 0. From the chain rule for the derivative of a composite function, we find that a composition of two analytic functions is analytic. More precisely, suppose that a function f(z) is analytic in a domain D and that the image of D under the transformation w = f(z) is contained in the domain of definition of a function g(w). Then the composition g[f(z)] is analytic in D, with derivative d dz g[f(z)] = g [f(z)]f (z) Theorem 2.. If f (z) = 0 everywhere in a domain D, then f(z) must be constant throughout D. Proof. Let f(z) = u(x, y)+iv(x, y). Assume that f (z) = 0 in D. since f is differentiable everywhere in D, the auchy-riemann conditions are satisfied throughout D. Since f (z) = 0 then u x + iv x = 0 and hence, v y iu y = 0. onsequently, u x = u y = v x = v y = 0 at each point in D. Let u(x, y) = g(y) and v(x, y) = h(y). Since u x = v y = 0 then g (y) = 0 and g(y) = c. Also, since v x = u y = 0 it follows that h (y) = 0 and h(y) = c 2. We then have f(z) = c + ic 2. Example 2.8. Determine the singular points of the function f(z) = z 3 + 4 (z 2 3)(z 2 + ) and state why the function is analytic everywhere except at those points. Solution. Since f(z) is a quotient of two polynomials, it is analytic in any domain throughout where the denominator (z 2 3)(z 2 + ) 0. Hence, the singular points are z = ± 3 and z = ±i.
46 HAPTER 2. ANALYTI FUNTIONS Example 2.9. Show that the function f(z) = cosh x cos y + i sinh x sin y is entire. Solution. Since f(z) = cosh x cos y + i sinh xsiny, we have Because u(x, y) = cosh x cos y and v(x, y) = sinh x sin y. u x = sinh x cos y = v y and u y = cosh x sin y = v x everywhere, it is clear from Theorem 2.9 that f is entire. Example 2.20. Suppose and its conjugate f(z) = u(x, y) + iv(x, y) f(z) = u(x, y) iv(x, y) are both analytic in a given domain D. Show that f(z) is constant in D. Solution. Let f(z) = U(x, y) + iv (x, y) where Since f(z) is analytic it follows that Also, since f(z) is analytic, U(x, y) = u(x, y) and V (x, y) = v(x, y). u x = v y and u y = v x. (2.9) U x = V y and U y = V x. which implies that u x = v y and u y = v x. (2.0) Adding corresponding sides of the first of Equations 2.9 and 2.9, we find that u x = 0. Similarly subtraction involving corresponding sides of the second of Equations 2.9 and 2.9 reveals that v x = 0. This implies that f (z) = u x + iv x = 0 + i0 = 0 and it follows from Theorem 2. that f(z) is constant throughout D.
2.. HARMONI FUNTIONS 47 Exercises 2.4.. Use Theorem 2.9 to verify that each of these functions is entire: (a) f(z) = 3x + y + i(3y x); (b) f(z) = sin x cosh y + i cos x sinh y; (c) f(z) = e y sin x ie y cos x; (d) f(z) = (z 2 2)e x e iy 2. Use Theorem 2.8 to show thateach of these functions is nowhere analytic: (a) f(z) = xy + iy; (b) f(z) = 2xy + i(x 2 y 2 (c) ); f(z) = e y e ix 3. In each case, determine the singular points of the function and state why the function is analytic everywhere except at those points: (a) f(z) = 2z + z(z 2 + ) ; (b) f(z) = z 3 + i z 2 3z + 2 ; (c) f(z) = z 2 + (z + 2)(z 2 + 2z + 2) 2. Harmonic Functions Definition 2.7. Let u be a real valued function of the real variables x and y. We say that u is harmonic in a given domain of R 2, if throughout that domain, it has continuous partial derivatives of the first and second order and it satisfies the Laplace s equation u xx + u yy = 0. Theorem 2.2. If a function f(z) = u(x, y) + iv(x, y) is analytic in a domain D, then its component functions u and v are harmonic in D. Proof. Assuming that f is analytic in D, we start with the observation that the first order partial derivatives of its component functions must satisfy the auchy-riemann equations throughout D: u x = v y and u y = v x. Differentiating both sides of these equations with respect to x, we have Likewise, differentiation with respect to y yields u xx = v yx and u yx = v xx. (2.) u xy = v yy and u yy = v xy. (2.2) Since the second order partial derivatives are continuous, u yx = u xy and v yx = v xy. It then follows from Equations 2. and 2.2 that Therefore, u and v are harmonic in D. u xx + u yy = 0 and v xx + v yy = 0.
48 HAPTER 2. ANALYTI FUNTIONS Definition 2.8. Let u and v be harmonic functions in a domain D. Moreover, if their first order partial derivatives satisfy the auchy-riemann equations throughout D, then v is said to be a harmonic conjugate of u. Theorem 2.3. A function f(z) = u(x, y) + iv(x, y) is analytic in a domain D if and only if v is a harmonic conjugate of u. Proof. If v is a harmonic conjugate of u in D, Theorem 2.9 tells us that f is analytic in D. onversely, if f is analytic in D, we know from Theorem 2.2 that u and v are harmonic in D; and, in view of Theorem 2.8, the auchy-riemann equations are satisfied. Example 2.2. Let u(x, y) = y 3 3x 2 y. Show that u is harmonic and find a harmonic conjugate v of u. Solution. First, we show that u is harmonic. u x = 6xy and u y = 3y 2 3x 2 u xx = 6y and u yy = 6y The partial derivatives are all continuous and u xx + u yy = 0. Thus, u is harmonic. We will now find a harmonic conjugate v of u such that u x = v y and u y = v x. Since u x = 6xy, it follows that v y = 6xy. v = ( 6xy) dy = 3xy 2 + g(x) Now, v x = 3y 2 + g (x). Since u y = v x, we have 3y 2 g (x) = 3y 2 3x 2 Hence, g (x) = 3x 2 and g(x) = x 3 +. Then the function is a harmonic conjugate of u(x, y). The corresponding analytic function is v(x, y) = 3xy 2 + x 3 + f(z) = y 3 3x 2 y + i( 3xy 2 + x 3 + )
2.. HARMONI FUNTIONS 49 Exercises 2.5.. Show that u(x, y) is harmonic in some domain and find a harmonic conjugate v(x, y) when (a) u(x, y) = 2x( y); (b) u(x, y) = 2x x 3 + 3xy 2 ; (c) u(x, y) = sinh x sin y; (d) u(x, y) = y/(x 2 + y 2 ). 2. Show that if v and V are harmonic conjugates of u in a domain D, thenv(x, y) and V (x, y) can differ at most by an additive constant. 3. Suppose that, in a domain D, a function v is a harmonic conjugate ofu and also that u is a harmonic conjugate of v. Show how it follows that both u(x, y) and v(x, y) must be constant throughout D.
50 HAPTER 2. ANALYTI FUNTIONS
hapter 3 ELEMENTARY FUNTIONS We consider here various elementary functions studied in calculus and define corresponding functions of a complex variable. To be specific, we define analytic functions of a complex variable z that reduce to the elementary, functions in calculus when z = x+i0. We start by defining the complex exponential function and then use it to develop the others. 3. The omplex Exponential Function Definition 3.. We define the complex exponential function f(z) = e z, by where y is in radians. f(z) = e z = e x+iy = e x e iy = e x (cos y + i sin y), We note that we can also write exp z instead of e z. Furthermore, Thus, f(z) = exp z is entire. d exp z = exp z, z. dz Theorem 3.. (Some Properties of the omplex Exponential Function) Let z = x + iy and z 2 = x 2 + iy 2. Then. (exp z )(exp z 2 ) = exp(z + z 2 ); 2. exp z exp z 2 = exp(z z 2 ); 3. The complex exponential function is periodic with period 2πi, i.e, 4. (exp z) n = exp nz; exp z = exp(z + 2πi). 5
52 HAPTER 3. ELEMENTARY FUNTIONS 5. exp z = exp x > 0. Thus, the function w = exp z maps the z-plane to the punctured w-plane, {0}. 6. arg exp z = y + 2πk Example 3.. Solve e z =. Solution. Let z = x + iy and e z =. Hence, e x e iy =. But = e iπ and e x e iy = e iπ. Therefore, z = i(2n + )π. Exercises 3... Show that e x = and y π = 2nπ e x = and y = π + 2nπ x = 0 and y = (2n + )π (a) exp(2 ± 3πi) = e 2 ; ( ) 2 + πi e (b) exp = ( + i); 4 2 (c) exp(z + πi) = exp z. 2. State why the function 2z 2 3 ze z + e z is entire. 3. Use the auchy-riemann equations and Theorem 2.8 to show that the function f(z) = exp z is not analytic anywhere. 4. Show in two ways that the function exp(z 2 ) is entire. What is its derivative? 5. Prove that exp( 2z) < if and only if Re z > 0. 6. Find all values of z such that (a) e z = 2; (b) e z = + 3 i; (c) exp(2z ) =. 7. Show that exp(iz) = exp(i z) if and only if z = nπ(n = 0, ±, ±2,...). 3.2 omplex Trigonometric Functions Definition 3.2. Let z. Then, cos z = eiz + e iz 2 and sin z = eiz e iz. 2i
3.2. OMPLEX TRIGONOMETRI FUNTIONS 53 Theorem 3.2. (Some Properties of omplex Trigonometric Functions) Let z = x + iy and z 2 = x 2 + iy 2. Then. d dz sin z = cos z and d dz cos z are entire. cos z = sin z. Moreover, both f(z) = sin z and f(z) = 2. The sine and cosine functions are both periodic with period 2π, i.e., sin(z + 2π) = sin z and cos(z + 2π) = cos z. 3. sin 2 z + cos 2 z = 4. sin(z +z 2 ) = sin z cos z 2 +cos z sin z 2 and cos(z +z 2 ) = cos z cos z 2 sin z sin z 2 5. sin z = sin x cosh y + i cos x sinh y and cos z = cos x cosh y i sin x sinh y 6. sin z 2 = sin 2 x + sinh 2 y and cos z 2 = cos 2 x + sinh 2 y. Example 3.2. Find z such that sin z = 0. Solution. We have sin z = 0 sin z = 0. Now, sin z 2 = 0 and sin 2 x + sinh 2 y = 0 which implies that This gives us Therefore, z = nπ. sin 2 x = 0 and sinh 2 y = 0. x = nπ and y = 0. Example 3.3. Show 2 sin z cos z 2 = sin(z + z 2 ) + sin(z l z 2 ). Solution. ( ) ( ) e iz e iz e iz 2 + e iz2 2 sin z cos z 2 = 2 2i 2 = ei(z+z2) e i(z2 z) + e i(z z2) e i(z+z2) 2i = ei(z+z2) e i(z+z2) + ei(z z2) e i(z+z2) 2i 2i = sin(z + z 2 ) + sin(z l z 2 ) The other four trigonometric functions are defined in terms of the sine and cosine functions by the usual relations: tan z = sin z cos z, cos z cot z = sin z, sec z = cos z, csc z = sin z
54 HAPTER 3. ELEMENTARY FUNTIONS Observe that the quotients tan z and sec z are analytic everywhere except at the singularities z = π + nπ, (n = 0, ±, ±2,...) 2 which are the zeros of cos z. Likewise, cot z and csc z have singularities at the zeros of sin z, namely z = nπ, (n = 0, ±, ±2,...). By differentiating the right-hand sides of the given equations, we obtain the expected differentiation formulas d dz tan z = sec2 z, d dz cot z = csc2 z Exercises 3.2. d dz sec z = sec z tan z, d dz csc z = csc z cot z.. Show that (a) + tan 2 z = sec 2 z; (b) + cot 2 z = csc 2 z. 2. Show that 2 sin(z + z 2 ) sin(z z 2 ) = cos 2z 2 cos 2z. 3. Use the auchy-riemann equations and Theorem 2.8 to show that neither sin z nor cos z is an analytic function of z anywhere. 4. Show that (a) cos(iz) = cos(i z) for all z; (b) sin(iz) = sin(i z) if and only if z = nπ(n = 0, ±, ±2,...). 3.3 omplex Hyperbolic Functions Definition 3.3. Let z. Then, sinh z = ez e z and cosh z = ez + e z. 2 2 Theorem 3.3. (Some Properties of omplex Hyperbolic Functions) Let z = x + iy and z 2 = x 2 + iy 2. Then. d sinh z = cosh z and d dz f(z) = cosh z are entire. dz cosh z = sinh z. Moreover, both f(z) = sinh z and 2. sin z = i sinh iz and cos z = cosh iz
3.3. OMPLEX HYPERBOLI FUNTIONS 55 3. sin iz = i sinh z and cos iz = cosh z 4. cosh( z) = cosh z and sinh( z) = sinh z; 5. cosh 2 z sinh 2 z = ; 6. sinh(z +z 2 ) = sinh z cosh z 2 +cosh z sinh z 2 and cosh(z +z 2 ) = cosh z cosh z 2 + sinh z sinh z 2 ; 7. sinh z = sinh z cos y + i cosh x sin y and cosh z = cosh x cos y + i sinh x sin y 8. sinh z 2 = sinh 2 x + sin 2 y The hyperbolic tangent of z is defined by the equation tanh z = sinh z cosh z and is analytic in every domain in which cosh z 0. The functions coth z, sechz, and cschz are the reciprocals of tanh z, cosh z, and sinh z, respectively. It is straightforward to verify the following differentiation formulas, which are the same as those established in calculus for the corresponding functions of a real variable: d dz tanh z = sech2 z, d dz coth z = csch2 z d dz sechz = sechz tan z, d cschz = cschz coth z. dz Exercises 3.3.. Prove that sinh 2z = 2 sinh z cosh z. 2. Show that (a) sinh(z + πi) = sinh z; (b) cosh(z + πi) = cosh z; (c) tanh(z + πi) = tanh z 3. Find all roots of the equation (a) sinh z = i; (b) cosh z = 2 ; 4. Find all roots of the equation cosh z = 2.
56 HAPTER 3. ELEMENTARY FUNTIONS 3.4 omplex Logarithmic Functions Recall that in R, the logarithmic function is the inverse of the exponential function. Having defined the exponential function as w = e z, we want to base our definition of the logarithmic function to the solution of the equation e w = z for w, where z 0. Definition 3.4. If z 0, we define log z by log z = ln r + i(θ + 2nπ), n Z. Example 3.4. If z = 3 i, find log z. Solution. Since z = ( 3 i), we have r = 2 and Θ = 2π/3. log z = log( 3 i) = ln 2 + i( 2π 3 + 2nπ) = ln 2 + 2(n 3 )π i We note that, there are many log z s, one for each arg z. Moreover, e log z = e ln r+iθ = e ln r e iθ = re iθ = z. This is similar to the logarithmic and exponential functions of a real variable. However, This is different from the real case. log(e z ) = z + 2nπi n Z. Definition 3.5. The principal value of the log z, denoted by Log z is defined as Log z = ln r + i Θ. We note that log z = Log z + 2nπi where n Z. Example 3.5. Find the principal value of log. Solution. Since z =, we have r = and Θ = 0. Log = ln + i(0) = 0 Example 3.6. Find the principal value of log( ). Solution. Since z =, we have r = and Θ = π. Log( ) = ln + i(π) = π i
3.4. OMPLEX LOGARITHMI FUNTIONS 57 If z = re iθ is a nonzero complex number, the argument θ has any one of the values θ = Θ + 2nπ, (n = 0, ±, ±2,...), where Θ = Arg z. Hence the definition can be written log z = ln r + i(θ + 2nπ) (3.) log z = ln r + iθ (3.2) If we let α denote any real number and restrict the value of θ so that α < θ < α+2π, the function log z = ln r + iθ is single-valued and continuous in the stated domain (Figure 3.). Figure 3.: Note that if the function log z = ln r+iθ were to be defined on the ray θ = α, it would not be continuous there. For, if z is a point on that ray, there are points arbitrarily close to z at which the values of v are near α and also points such that the values of v are near α + 2π, The function log z = ln r + iθ is not only continuous but also analytic in the domain r > 0, α < θ < α + 2π since the first-order partial derivatives of u and v are continuous there and satisfy the polar form of the auchy-riemann equations. Also, ru r = v θ, u θ = rv r d dz log z = z Definition 3.6. A branch of a multiple-valued function f is any single-valued function F that is analytic in some domain at each point z of which the value F (z) is one of the values f(z). The function Log z = ln r + i Θ, (r > 0, π < Θ < π) is called the principal branch.
58 HAPTER 3. ELEMENTARY FUNTIONS Definition 3.7. A branch cut is a portion of a line or curve that is introduced in order to define a branch F of a multiple-valued function f. Points on the branch cut for F are singular points of F, and any point that is common to all branch cuts of f is called a branch point. The origin and the ray θ = α make up the branch cut for the branch log z = ln r + iθ of the logarithmic function. The branch cut for the principal branch Log z = ln r + i Θ consists of the origin and the ray Θ = π. The origin is evidently a branch point for branches of the multiple-valued logarithmic function. Remark 3... log(e z ) = z + 2nπi, n Z; Log (e z ) = z; 2. log(z z 2 ) = log z + log z 2 ; ) 3. log ( z z 2 Exercises 3.4.. Show that = log z log z 2 (a) Log( ei) = π 2 i; (b) Log( i) = 2 ln 2 π 4 i 2. Verify that when n = 0, ±, ±2,..., (a) log e = + 2nπi; ( (b) log i = 2n + ) πi; 2 (c) log( + ( 3 i) = ln 2 + 2 n + ) πi 3 3. Show that (a) Log( + i) 2 = 2Log( + i); (b) Log( + i) 2 2Log( + i) 4. Show that (a) log(i 2 ) = 2 log i when log z = ln r + iθ (b) log(i 2 ) 2 log i when log z = ln r + iθ 5. Find all roots of the equation log z = iπ/2. ( ( r > 0, π 4 < θ < 9π 4 ) r > 0, 3π 4 < θ < π 4 ; )
3.5. OMPLEX INVERSE TRIGONOMETRI FUNTIONS 59 3.5 omplex Inverse Trigonometric Functions In order to define the inverse sine function sin z, we write Hence, If we put this equation in the form and solve for e iw, we find that w = sin z when z = sin w. z = eiw e iw. 2i (e iw ) 2 2iz(e iw ) = 0 e iw = iz + ( z 2 ) /2. Taking logarithms of each side of the equation and recalling that w = sin z, we have the definition of the inverse since function. Definition 3.8. The inverse sine function is defined as sin z = i log[iz + ( z 2 ) /2 ] Example 3.7. Find all values of sin ( i). Solution. From Definition 3.8, sin ( i) = i log[i( i) + ( ( i) 2 ) /2 ] = i log( ± 2). But and Since then log( + 2) = ln( + 2) + 2nπi (n = 0, ±, ±2,...) log( 2) = ln( 2 ) + (2n + )πi (n = 0, ±, ±2,...). ln( 2 ) = ln + 2 = ln( + 2) ( ) n ln( + 2) + nπi (n = 0, ±, ±2,...) constitute the set of values of log( + 2). Thus, in rectangular form, sin ( i) = nπ + i( ) n+ ln( + 2) (n = 0, ±, ±2,...). Definition 3.9. The inverse cosine function is defined as cos z = i log[z + i + ( z 2 ) /2 ]. The inverse tangent function is defined as tan z = i 2 log i + z i z.
60 HAPTER 3. ELEMENTARY FUNTIONS Remark 3.2. The derivatives of these three functions are given below.. 2. d dz sin z = ( z 2 ) /2; d dz cos z = ( z 2 ) /2; 3. d dz tan z = + z 2. Exercises 3.5.. Find all the values of (a) tan (2i); (b) tan ( + i) 2. Solve the equation sin z = 2 for z. 3. Solve the equation cos z = 2 for z. 4. Derive the formula for the derivative of sin z. 5. Derive the expression for tan z. 6. Derive the formula for the derivative of tan z.
hapter 4 INTEGRALS Integrals are extremely important in the study of functions of a complex variable. The theory of integration, to be developed in this chapter, is noted for its mathematical elegance. 4. Definite Integrals of w(t) In R, we have b f(x)dx, where f is continuous on a closed interval [a, b]. We want to a define z2 z f(z)dz, (4.) where f is a continuous complex valued function. Since intervals do not exist in, the integral (4.) will be taken over a curve which joins z and z 2. Definition 4.. Let w(t) be a complex-valued function of a real variable t written as w(t) = u(t) + iv(t), (4.2) where u and v are real-valued, the definite integral of w(t) over an interval a t b is defined as b a w(t)dt = b when the definite integral on the right exist. Thus, Re b a w(t)dt = b a a u(t)dt + i Re[w(t)]dt and Im 6 b a b a v(t)dt, (4.3) w(t)dt = b a Im[w(t)]dt
62 HAPTER 4. INTEGRALS Example 4.. Evaluate 0 ( + it) 2 dt. Solution. ( + it) 2 dt = Example 4.2. Evaluate Solution. 0 π/4 0 π/4 0 e it dt. 0 ( t 2 )dt + i e it dt = ie it] π/4 0 0 2tdt = 2 3 + i = ie iπ/4 + i ( = i 2 + i ) + i 2 = ( + i ) 2 2 == Theorem 4.. Let w = u + iv, W = U + iv.. 2. 3. 4. b b z 0 w(t)dt = z 0 w(t)dt. a a b [w(t) + W (t)]dt = b w(t)dt + b a a a b a a b a w(t)dt = w(t)dt b b a w(t)dt w(t) dt W (t)dt Exercises 4... Evaluate the following integrals: (a) 2 ( ) 2 t i dt; (b) π/6 e i2t dt; 2. Show that if m and n are integers, 2π 0 e imθ e inθ dθ = { 0, when m n; 2π, when m = n.
4.2. ARS AND ONTOURS 63 3. Given π e (+i)x dx = 0 π 0 e x cos x dx + i π 0 e x sin x dx. Evaluate the two integrals on the right by evaluating the single integral on the left and then using the real and imaginary parts of the value found. 4. For n Z +, for all x [, ], Show that P n (x). P n (x) = π π 0 (x + i x 2 cos θ) n dθ. 4.2 Arcs and ontours Integrals of complex-valued functions of a complex variable are defined on curves in the complex plane, rather than on just intervals of the real line. lasses of curves that are adequate for the study of such integrals are introduced in this section. Definition 4.2. An arc on the complex plane is a set of points z = (x, y) such that x = x(t); y = y(t), t [a, b], where x(t) and y(t) are continuous on [a, b], that is is dependent on t and we may write = {z = z(t) = x(t) + iy(t); t [a, b]}. The equation z = z(t) is called the parametric representation of the arc. The arc is a simple arc, or a Jordan arc, if it does not cross itself that is, is simple if z(t l ) z(t 2 ) when t l t 2. When the arc is simple except for the fact that z(b) = z(a), we say that is a simple closed curve, or a Jordan curve. Figure 4.:
64 HAPTER 4. INTEGRALS Example 4.3. Draw the polygonal line defined by means of the equations z = { x + ix, when 0 x ; x + i, when x 2. to illustrate a simple arc. Solution. Figure 4.2: Example 4.4. Verify that the unit circle z = e iθ, 0 θ 2π about the origin is a simple closed curve oriented in the counterclockwise direction. Solution. Figure 4.3: Example 4.5. Verify that the unit circle z = e iθ, 0 θ 2π about the origin is a simple closed curve oriented in the clockwise direction.
4.2. ARS AND ONTOURS 65 Solution. Figure 4.4: The arcs in Figures 4.3 and 4.4 are different. The set of points is the same, however, the former is traversed in the counterclockwise direction while the latter is traversed in the clockwise direction. Figure 4.5: The points on the arc z = e i2θ, 0 θ 2π (see Figure 4.5) are the same as those making up the arcs in Figures 4.3 and 4.4. The arc here differs, however, from each of those arcs since the circle is traversed twice in the counterclockwise direction. Remark 4.. A closed curve is positive oriented if it is traversed in the counterclockwise direction. Definition 4.3. We define the following terms.. If an arc is defined by z(t) = x(t) + iy(t), then the arc is said to be differentiable if x (t) and y (t) are continuous on [a, b]. If z (t) = x (t)+iy (t), then z (t) represents the tangent vector to the arc for a given t. 2. If an arc is differentiable, then L(t) = z (t) = [x (t)] 2 + [y (t)] 2,
66 HAPTER 4. INTEGRALS is continuous on [a, b], then we say that L(t) is integrable on [a, b]. We note that the expression b [x (t)] 2 + [y (t)] 2 dt, is the length of arc. a 3. Given an arc, if z (t) is continuous for all t [a, b] and z (t) 0, for all t (a, b), then is said to be a smooth arc in [a, b]. 4. An arc is said to be piecewise smooth or a contour if it is the union of a finite number of smooth arcs joined end to end. Exercises 4.2.. Suppose that a function f(z) is analytic at a point z 0 = z(t 0 ) lying on a smooth arc z = z(t)(a t b). Show that if w(t) = f[z(t)],then when t = t 0. w (t) = f [z(t)]z (t) 2. Let y(x) be a real-valued function defined on the interval 0 x by means of the equations ( ) π x 3 sin, when 0 < x ; x 0, when x = 0. Show that the equation z = x + iy(x) (0 x ) represents an arc that intersects the real axis at the points z = /n(n =, 2,...) and z = 0, as shown in Figure 4.6. Figure 4.6:
4.3. ONTOUR INTEGRALS 67 4.3 ontour Integrals We turn now to integrals of complex-valued functions f of the complex variable z. Such an integral is defined in terms of the values f(z) along a given contour, extending from a point z = z to a point z = z 2 in the complex plane. Definition 4.4. Let be a contour parametrized by z = z(t), t [a, b], and let f be a complex valued function continuous on the. The contour integral or the line integral of f over, denoted by f(z)dz is defined by Example 4.6. Evaluate f(z)dz = b a f(z(t))z (t)dt. z dz when is the right-hand half of the circle z = 2, from z = 2i to z = 2i. Solution. Figure 4.7: We have z = 2e it, π 2 t π 2. Hence, z dz = = = 4i π/2 π/2 π/2 = 4πi π/2 π/2 π/2 f(z(t))z (t)dt 2e it 2ie it dt dt
68 HAPTER 4. INTEGRALS Theorem 4.2. The following are some properties of contour integrals.. If = + 2, then f(z)dz = f(z)dz + f(z)dz. 2 2. If z 0, then z 0 f(z)dz = z 0 f(z)dz. 3. Suppose is a contour defined by z = z(t), t [a, b], then is the contour containing the same points as but the path is traced in the reversed order. Thus, is parametrized by z = z( t), t [ b, a]. Furthermore, f(z)dz = f(z)dz. 4. [f(z) ± g(z)]dz = f(z)dz ± g(z)dz 5. If f(z) M, z, and has length L, then f(z)dz ML. Example 4.7. Let be the contour OAB shown in Figure 4.8 and evaluate the integral f(z)dz = OA f(z)dz + f(z)dz AB where f(z) = y x i3x 2, (z = x + iy).
4.3. ONTOUR INTEGRALS 69 Figure 4.8: Solution. The leg OA may be represented parametrically as z = 0 + iy, 0 y ; and since x = 0 at points on that leg, the values of f there vary with the parameter y according to the equation f(z) = y, 0 y. onsequently, OA f(z)dz = On the leg AB, z = x + i, 0 x ; and so AB f(z)dz = 0 ( x i3x 2 ) dx = 0 yi dy = i 0 0 y dy = i 2 ( x) dx 3i 0 x 2 dx = 2 i Hence, f(z)dz = i 2 If 2 denotes the segment OB of the line y = x, with parametric representation z = x + ix, 0 x, 2 f(z)dz = 0 ( i3x 2 )( + i) dx = 3( i) 0 x 2 dx = i Evidently, then, the integrals of f(z) along the two paths and 2 have different values even though those paths have the same initial and the same final points. Observe how it follows that the integral of f(z) over the simple closed contour OABO, or l 2, has the nonzero value f(z)dz f(z)dz = + i. 2 2
70 HAPTER 4. INTEGRALS Example 4.8. Evaluate when is the semicircle z = 3e iθ, 0 θ π Solution. z /2 dz Figure 4.9: We observe that when z(θ) = 3e i0, the right-hand limits of the real and imaginary components of the function f[z(θ)] = 3e iθ/2 = 3 cos θ 2 + i 3 cos θ 2, 0 θ π at θ = 0 are 3 and 0, respectively. Hence f[z(θ)] is continuous on the closed interval 0 θ π when its value at θ = 0 is defined as 3. onsequently, z /2 dz = π 0 3e iθ/2 3ie iθ dθ π = 3 3 i e i3θ/2 dθ 0 = 3 [ ] π 2 3 i = 3 3 i 3i ei3θ/2 = 2 3( + i) 0 [ 23i ( + i) ] Exercises 4.3. For the functions f and contours in Exercises through 5, use parametric representations for, or legs of, to evaluate f(z)dz.
4.4. ANTIDERIVATIVES 7. f(z) = z + 2 z and is (a) the semicircle z = 2e iθ, 0 θ π; (b) the semicircle z = 2e iθ, π θ 2π; (c) the circle z = 2e iθ, 0 θ 2π. 2. f(z) = z and is the arc from z = 0 to z = 2 consisting of (a) the semicircle z = + e iθ, π θ 2π; (b) the segment 0 x 2 of the real axis. 3. f(z) = π exp(π z) and is the boundary of the square with vertices at the points 0,, + i, and i, the orientation of being in the counterclockwise direction. 4. f(z) is defined by the equations f(z) = {, when y < 0; 4y, when y > 0. and is the arc from z = i to z = + i along the curve y = x 3. 5. f(z) = and is an arbitrary contour from any fixed point z to any fixed point z 2 in the plane. 6. Evaluate using this representation for : z dz z = 4 y 2 + iy, 2 y 2. 7. Let 0 denote the circle z z 0 = R, taken counterclockwise. Use the parametric representation z = z 0 +Re iθ ( π θ π) for 0 to derive the following integration formulas: (a) (b) 0 dz z z 0 = 2πi; 0 (z z 0 ) n dz = 0, n = ±, ±2,.... 4.4 Antiderivatives Definition 4.5. We say F is an antiderivative of f in some domain D if F (z) = f(z), for all z in D.
72 HAPTER 4. INTEGRALS Theorem 4.3. Suppose that a function f(z) is continuous on a domain D. The following statements are equivalent:. f(z) has an antiderivative F (z) in D; 2. the integrals of f(z) along contours lying entirely in D and extending from any fixed point z to any fixed point z 2 all have the same value; 3. the integrals of f(z) around closed contours lying entirely in D all have value zero. Proof. () (2) Suppose f(z) has an antiderivative F (z) in D; i.e., F (z) = f(z). Let z, z 2 D and consider z2 z parametric representation z = z(t), a t b. Now, f(z)dz. hoose a contour from z to z 2, lying in D, with f(z)dz = b a f(z(t))z (t)dt = F (z(t)) b a = F (z(b)) F (z(a)) = F (z 2 ) F (z ) Thus the integrals of f(z) are path independent. In this case, we write z2 z = F (z 2 ) F (z ). (2) (3) We let z and z 2 denote any two points on a closed contour lying in D and form two paths, each with initial point z and final point z 2, such that = 2 (Figure 4.0). Assuming that the integrals of f(z) along contours lying entirely in D and extending from any fixed point z to any fixed point z 2 all have the same value, we have f(z)dz = f(z)dz 2 or f(z)dz = f(z)dz = 0. 2 That is, the integral of f(z) around the closed contour = 2 has value zero.
4.4. ANTIDERIVATIVES 73 Figure 4.0: (3) (2) We let and 2 denote any two contours, lying in D, from a point z l to a point z 2. From (3), the integrals of f(z) around closed contours lying entirely in D all have value zero. Hence, 0 = f(z)dz = f(z)dz 2 = f(z)dz + f(z)dz 2 = f(z)dz f(z)dz 2 Therefore, f(z)dz f(z)dz. 2 (2) () We want to find F (z) such that F (z) = f(z). hoose z 0 D. Define F (z) = z z 0 f(s)ds on D. Let z + z be any point, distinct from z, lying in some neighborhood of z that is small enough to be contained in D. Then F (z + z) F (z) = = = z+ z z 0 z z 0 f(s)ds + z+ z z z f(s)ds f(s)ds z 0 f(s)ds z+ z z f(s)ds z z 0 f(s)ds
74 HAPTER 4. INTEGRALS where the path of integration from z to z + z may be selected as a line segment (Figure 4.). Since + z+ z z ds = z, we can write f(z) = z z+ z z f(z)ds and it follows that F (z + z) F (z) f(z) = z z z+ z z [f(s) f(z)]ds. But f is continuous at the point z. Hence, for each positive number ε, a positive number δ exists such that f(s) f(z) < ε whenever s z < δ. onsequently, if the point z + z is close enough to z so that z < δ, then F (z + z) F (z) z f(z) < ε z = ε; z that is, F (z + z) F (z) lim = f(z), z 0 z or F (z) = f(z) Figure 4.:
4.4. ANTIDERIVATIVES 75 Example 4.9. Evaluate z 2 dz where is a contour that starts at z = 0 and ends at z = + i. Solution. We observe that F (z) = z3 3, F (z) = f(z). Hence f has an antiderivative and z 2 dz = +i 0 +i = z3 = 3 0 ( + i)3 3 z 2 dz 0 = 2 ( + i) 3 Example 4.0. Evaluate dz z where is the right half of the circle z = 2e iθ, π/2 θ π/2 from z = 2i to z 2 = 2i. Solution. Figure 4.2:
76 HAPTER 4. INTEGRALS Recall that d dz [Log z] = z. dz z = 2i 2i dz z = Log(2i) Log( 2i) ( ) ( = ln 2 + i π 2 ln 2 i π 2 ) Example 4.. Evaluate = πi 2 where 2 is the left half of the circle z = 2e iθ, π/2 θ 3π/2 from z = 2i to z 2 = 2i. Solution. dz z Figure 4.3: 2 dz z = 2i 2i dz z = log( 2i) log(2i) ( ) ( = ln 2 + i 3π 2 ln 2 + i π 2 ) = πi From Examples 4.0 and 4., the value of the integral of /z around the entire circle = + 2 is thus obtained: dz z = dz z + dz = πi + πi = 2πi 2 z
4.5. AUHY-GOURSAT THEOREM 77 Exercises 4.4.. Use an antiderivative to show that, for every contour extending from a point z l to a point z 2, z n dz = n + (zn+ 2 z n+ ), n = 0,, 2,.... 2. By finding an antiderivative, evaluate each of these integrals, where the path is any contour between the indicated limits of integration: (a) ) i/2 e πz dz; (c) 3 i (z 2)3 dz. 3. Use Theorem 4.3 to show that (b) π+2i 0 cos ( z 2 dz; 0 (z z 0 ) n dz = 0, n = ±, ±2,.... when 0 is any closed contour which does not pass through the point z 0. 4.5 auchy-goursat Theorem In the previous section, we saw that when a continuous function f has an antiderivative in a domain D, the integral of f(z) around any given closed contour lying entirely in D has value zero. In this section, we present a theorem giving other conditions on a function f, which ensure that the value of the integral of f(z) around a simple closed contour is zero. We let denote a simple closed contour z = z(t), a t b), described in the positive sense (counterclockwise), and we assume that f is analytic at each point interior to and on. We have b f(z)dz = f(z(t))z (t)dt and if a f(z) = u(x, y) + iv(x, y) and z(t) = x(t) + iy(t), the integrand f(z(t))z (t) is the product of the functions of the real variable t. Thus f(z)dz = u[x(t), y(t)] + iv[x(t), y(t)], x (t) + iy (t) b a (ux vy )dt + i(vx + uy )dt. In terms of line integrals of real-valued functions of two real variables, then, f(z)dz = u dx v dy + i v dx + u dy. (4.4)
78 HAPTER 4. INTEGRALS Theorem 4.4. (Green s Theorem) Suppose that two real-valued functions P (x, y) and Q(x, y), together with their first-order partial derivatives, are continuous throughout the closed region R consisting of all points interior to and on the simple closed contour. Then P dx + Qdy = (Q x P y )da. Now f is continuous in R, since it is analytic there. Hence the functions u and v are also continuous in R. Likewise, if the derivative f of f is continuous in R, so are the first-order partial derivatives of u and v. Green s theorem then enables us to rewrite Equation 4.4 as f(z)dz = ( v x u y )da + i (u x v y )da. R R But, in view of the auchy-riemann equations, R u x = v y, u y = v x the integrands of these two double integrals are zero throughout R. analytic in R and f is continuous there, f(z)dz = 0. So, when f is This result was obtained by auchy in the early part of the nineteenth century. To illustrate, if is any simple closed contour, in either direction, then exp(z 3 )dz = 0. This is because the function f(z) = exp(z 3 ) is analytic everywhere and its derivative f (z) = 3z 2 exp(z 3 ) is continuous everywhere. Goursat was the first to prove that the condition of continuity on f can be omitted. Its removal is important and will allow us to show, for example, that the derivative f of an analytic function f is analytic without having to assume the continuity of f, which follows as a consequence. We now state the revised form of auchy s result, known as the auchy-goursat Theorem. Theorem 4.5. (auchy-goursat Theorem) If a function f is analytic at all points interior to and on a simple closed contour, then f(z)dz = 0. We need the following lemma to prove the auchy-goursat Theorem. We start by forming subsets of the region R which consists of the points on a positively oriented simple closed contour together with the points interior to. To do this, we draw equally spaced lines parallel to the real and imaginary axes such that the distance between adjacent vertical lines is the same as that between adjacent horizontal lines.
4.5. AUHY-GOURSAT THEOREM 79 We thus form a finite number of closed square subregions, where each point of R lies in at least one such subregion and each subregion contains points of R. We refer to these square subregions simply as squares, always keeping in mind that by a square we mean a boundary together with the points interior to it. If a particular square contains points that are not in R, we remove those points and call what remains a partial square. We thus cover the region R with a finite number of squares and partial squares (Figure 4.4) and our proof of the following lemma starts with this covering. Figure 4.4: Lemma 4.. Let f be analytic throughout a closed region R consisting of the points interior to a positively oriented simple closed contour together with the points on itself. For any positive number ε, the region R can be covered with a finite number of squares and partial squares, indexed by j =, 2,..., n, such that in each one there is a fixed point z for which the inequality f(z) f(z j ) f (z j ) z z j < ε, (z z j) (4.5) is satisfied by all other points in that square or partial square. Proof. To start the proof, we consider the possibility that, in the covering constructed just prior to the statement of the lemma, there is some square or partial square in which no point z j exists such that Inequality 4.5 holds for all other points z in it. If that subregion is a square, we construct four smaller squares by drawing line segments joining the midpoints of its opposite sides (Figure 4.4). If the subregion is a partial square, we treat the whole square in the same manner and then let the portions that lie outside R be discarded. If, in any one of these smaller subregions, no point z exists such that Inequality 4.5 holds for all other points z in it, we construct still smaller squares and
80 HAPTER 4. INTEGRALS partial squares, etc. When this is done to each of the original subregions that requires it, it turns out that, after a finite number of steps, the region R can be covered with a finite number of squares and partial squares such that the lemma is true. To verify this, we suppose that the needed points z do not exist after subdividing one of the original subregions a finite number of times and reach a contradiction. We let σ 0 denote that subregion if it is a square; if it is a partial square, we let σ 0 denote the entire square of which it is a part. After we subdivide σ 0, at least one of the four smaller squares, denoted by σ, must contain points of R but no appropriate point z j. We then subdivide σ and continue in this manner. It may be that after a square σ k, k =, 2,... has been subdivided, more than one of the four smaller squares constructed from it can be chosen. To make a specific choice, we take σ k to be the one lowest and then furthest to the left. In view of the manner in which the nested infinite sequence σ 0, σ, σ 2,..., σ k, σ k,... of squares is constructed, it can be shown that there is a point z 0 common to each σ k ; also, each of these squares contains points R other than possibly z 0. Recall how the sizes of the squares in the sequence are decreasing, and note that any δ neighborhood z z 0 < δ of z 0 contains such squares when their diagonals have lengths less than δ. Every δ neighborhood z z 0 < δ therefore contains points of R distinct from z 0, and this means that z 0 is an accumulation point of R. Since the region R is a closed set, it follows that z 0 is a point in R. Now the function f is analytic throughout R and, in particular, at z 0. onsequently, f (z 0 ) exists. According to the definition of derivative, there is, for each positive number ε, a δ neighborhood z z 0 < δ such that the inequality f(z) f(z 0 ) f (z 0 ) z z 0 < ε (4.6) is satisfied by all points distinct from z 0 in that neighborhood. But the neighborhood z z 0 < δ contains a square σ K when the integer K is large enough that the length of a diagonal of that square is less than δ (Figure 4.5). onsequently, z 0 serves as the point z j in Inequality 4.5 for the subregion consisting of the square σ K or a part of σ K. ontrary to the way in which the sequence σ 0, σ, σ 2,..., σ k, σ k,... was formed, then, it is not necessary to subdivide σ K. We thus arrive at a contradiction, and the proof of the lemma is complete.
4.5. AUHY-GOURSAT THEOREM 8 Figure 4.5: We will now prove the auchy-goursat Theorem. Proof. Let f be a function which is analytic throughout a region R consisting of a positively oriented simple closed contour and points interior to it. Given an arbitrary positive number ε, we consider the covering of R in the statement of the lemma. Let us define on the jth square or partial square the following function, where z is the fixed point in that subregion for which Inequality 4.5 holds: f(z) f(z j ), when z z δ j (z) = j, z z j (4.7) 0, when z = z j. According to Inequality 4.5, δ j (z) < ε at all points z in the subregion on which δ j (z) is defined. Also, the function δ j (z) is continuous throughout the subregion since f(z) is continuous there and lim z z j δ j (z) = f (z j ) f (z j ) = O. Next, let j, (j =, 2,..., n) denote the positively oriented boundaries of the above squares or partial squares covering R. In view of 4.7, the value of f at a point z on any particular j can be written f(z) = f(z j ) z j f (z j ) + f (z j )z + (z z j )δ j (z) and this means that f(z)dz = [f(z j ) z j f (z j )] j dz + f (z j ) j z dz + j (z z j )δ j (z)dz j
82 HAPTER 4. INTEGRALS But j dz = 0 and j z dz = 0 since the functions and z possess antiderivatives everywhere in the finite plane. So f(z)dz = (z z j )δ j (z)dz, j =, 2,..., n. j j The sum of all n integrals on the left can be written Σ j= f(z)dz = f(z)dz j since the two integrals along the common boundary of every pair of adjacent subregions cancel each other, the integral being taken in one sense along that line segment in one subregion and in the opposite sense in the other (Figure 4.6). Only the integrals along the arcs that are parts of remain. Thus, f(z)dz = Σ n j= (z z j )δ j (z)dz; j and so f(z)dz Σn j= (z z j )δ j (z)dz j. Now, recall that each j coincides either entirely or partially with the boundary of a square. In either case, we let s j denote the length of a side of the square. Since, in the jth integral, both the variable z and the point z j lie in that square, z z j 2 s j. Since δ j (z) < ε, then, we know that each integrand on the right satisfies the condition z z j δ j (z) 2 s j ε. As for the length of the path j, it is 4s j if j is the boundary of a square. In that case, we let A j denote the area of the square and observe that z z j δ j (z)dz j < 2 s j ε4s j = 4 2A j ε. If j is the boundary of a partial square, its length does not exceed 4s j + L j, where L j is the length of that part of j which is also a part of. Again letting A j denote the area of the full square, we find that z z j δ j (z)dz j < 2 s j ε(4s j + L j ) < 4 2A j ε + 2SL j ε,
4.5. AUHY-GOURSAT THEOREM 83 where S is the length of a side of some square that encloses the entire contour as well as all of the squares originally used in covering R (Figure 4.6). Note that the sum of all the A j s does not exceed S 2. If L denotes the length of, it now follows that f(z)dz < (4 2S 2 + 2 SL)ε Since the value of the positive number ε is arbitrary, we can choose it so that the righthand side of this last inequality is as small as we please. The left-hand side, which is independent of ε, must therefore be equal to zero; and f(z)dz = 0. Figure 4.6: Definition 4.6. A domain D is simply connected if every simple closed contour in D encloses only elements of D, that is the interior of is a subset of D for every simple closed contour which lies in D. A domain which is not simply connected is said to be multiply-connected.
84 HAPTER 4. INTEGRALS Figure 4.7: The auchy-goursat Theorem can be extended in the following way, involving a simply connected domain. Theorem 4.6. If a function f is analytic throughout a simply connected domain D, then f(z)dz = 0 for every closed contour lying in D. Proof. Let be a simple closed contour or a closed contour that intersects itself a finite number of times. Since is simple and lies in D, the function f is analytic at each point interior to and on ; and the auchy-goursat theorem ensures that f(z)dz = 0 holds. Furthermore, if is closed but intersects itself a finite number of times, it consists of a finite number of simple closed contours. This is illustrated in Figure 4.8, where the simple closed contours k (k =, 2, 3, 4) make up. Since the value of the integral around each k is zero, according to the auchy-goursat theorem, it follows that f(z)dz = 4 k= k f(z)dz = 0