Chapter 6 Complex Function Contents 6. Complex Number 3 6.2 Elementary Functions 6.3 Function of Complex Variables, Limit and Derivatives 3 6.4 Analytic Functions and Their Derivatives 8 6.5 Line Integral in the Complex Plane 2 6.6 Cauchy s Integral Theorem and Formula 23 6.7 Sequence and Series 25 6.8 Laurent s Theorem 29 6.9 Singularity and Zeros, Infinity 30 6.0 Residue Integration Method 3 6. Exercise 34 Complex function is the branch of mathematical analysis that investigates functions of complex numbers. It is useful in many branches of mathematics including algebraic geometry, number theory, applied mathematics as well as in physics including hydrodynamics and thermodynamics and also in engineering fields such as nuclear, aerospace, mechanical and electrical engineering. 6. Complex Number Definition 6.. An expression of the form z = x + iy, where x and y are real numbers and i = or i 2 = is called a complex number and the set of complex number denoted by C = {x + iy x,y R and i = } Example 6. z = 5 + i3 and z 2 = + i0 are particular examples of complex numbers. In the expression of a complex number z = x + iy, x is called the real part; that is, Re z = x and y is called the imaginary part of the complex number; that is, Im z = y. Example 6.2 z = 4 3i is an example of complex number in which 4 is a real part and -3 is an imaginary part. Definition 6..2 Two complex numbers z = x + iy and z 2 = x 2 + iy 2 are said to be equal; that is, z = z 2 if and only if x = x 2 and y = y 2.
2 Complex Function Figure 6.: z = x + iy Figure 6.2: z = 4 3i Definition 6..3 The sum of two complex numbers z = x + iy and z 2 = x 2 + iy 2 is defined by z + z 2 = (x + iy ) + (x 2 + iy 2 ) = (x + x 2 ) + i(y + y 2 ) Figure 6.3: The sum of two complex number Definition 6..4 The difference of two complex numbers z = x + iy and z 2 = x 2 + iy 2 is defined by z z 2 = (x x 2 ) + i(y y 2 ) is again a complex number. Properties of addition for the set of complex number Let z = x + iy, z 2 = x 2 + iy 2 and z 3 = x 3 + iy 3 are any complex numbers, then. Closure Properties z + z 2 C 2. Commutative Properties z + z 2 = z 2 + z
6. Complex Number 3 Figure 6.4: The difference of two complex number 3. Associative Properties 4. Scaler Multiplication For any z = x + iy in C and α any constant (z + z 2 ) + z 3 = z + (z 2 + z 3 ) αz = αx + iαy C 5. Existence of Identity For any z = x + iy in C, there exist an element 0 = 0 + i0 in C such that z + 0 = (x + iy) + (0 + i0) = x + iy = (0 + i0) + (x + iy) = 0 + z Hence we call 0 + i0 (denoted by 0) the additive identity element on C. 6. Existence of Inverse For any z = x + iy in C, there exist an element z = x iy in C such that z + ( z) = (x + iy) + ( x iy) = 0 + i0 = ( x iy) + (x + iy) = ( z) + z Hence z = x iy is called the additive inverse of element on C Example 6.3 Evaluate (2 + i) + (5 + 3i). Solution: (2 + i) + (5 + 3i) = (2 + 5) + (i + 3i) = 7 + 4i Example 6.4 Evaluate 2(3 + 5i) 3(2 i). Solution: 2(3 + 5i) 3(2 i) = (6 + 0i) (6 3i) = (6 6) + (0i + 3i) = 3i
4 Complex Function Definition 6..5 The product of two complex numbers z = x + iy and z 2 = x 2 + iy 2 is defined by z z 2 = (x + iy ) (x 2 + iy 2 ) = (x x 2 y y 2 ) + i(x y 2 + x 2 y ) is again a complex number. Properties of Multiplication for the set of complex number Let z = x + iy, z 2 = x 2 + iy 2 and z 3 = x 3 + iy 3 are any complex numbers, then. Closure Properties z z 2 C 2. Commutative Properties 3. Associative Properties 4. Distributive Properties z z 2 = z 2 z (z z 2 ) z 3 = z (z 2 z 3 ) z (z 2 + z 3 ) = (z z 2 ) + (z z 3 ) 5. Existence of Identity For any z = x + iy in C, there exist an element = + i0 in C such that z = (x + iy) ( + i0) = x + iy = ( + 0i) (x + iy) = z Hence we call = + i0 the multiplication identity element on C. 6. Existence of Inverse x iy For any z = x + iy 0 in C, there exist an element in C such that x 2 +y 2 (x + iy)( x iy x 2 + y 2 ) = = ( x iy x 2 )(x + iy) + y2 Hence x iy x 2 +y 2 is called the multiplicative inverse of element on C. Conjugate and Module of Complex Number Definition 6..6 Given a complex number z = x + iy. The conjugate of z, denoted by z, is defined by z = x iy. Example 6.5 The complex conjugate of z = 5 + 2i is z = 5 2i Figure 6.5: z = 5 2i Remark 6... Every real number has its own conjugate. 2. The product of any complex number with its conjugate is a real number; that is, if z = x + iy is a
6. Complex Number 5 complex number, then z z = (x + iy) (x iy) = x 2 i(xy) + i(xy) i 2 y 2 = x 2 ( )y 2 = x 2 + y 2 here x 2 + y 2 is areal number Properties of Conjugate. For any complex number z = x + iy, we have the following properties (a) z = z (b) z + z = 2x (c) z z = 2iy 2. Let z = x + iy and z 2 = x 2 + iy 2 be any two complex numbers. Then (a) z + z 2 = z + z 2 (b) z z 2 = z z 2 (c) z z 2 = z z 2 (d) ( z z2 ) = z2 z Definition 6..7 Consider the complex number z = x + iy. The modules(or absolute value) of z, denoted by z, is defined by z = x 2 + y 2 Properties of Modules. For any complex number z, z > 0. But if z = 0, then z = 0. 2. Consider the complex number z = x + iy. Since z = x iy, we have z = x 2 + ( y) 2 = x 2 + y 2 = z 3. z = z 4. z z = x 2 + y 2 = z 2 for any complex number z. 5. z z 2 = z z 2 6. z z2 = z z 2 7. z + z 2 2 = z 2 + 2Re(z z 2 ) + z 2 2 8. z z 2 2 = z 2 2Re(z z 2 ) + z 2 2 Example 6.6 Let z = 2 3i. Then find z and z. Solution: Since z = 2 i3 we have z = 2 + 3i, then z = 2 2 + ( 3) 2 = 3 = z z z = (2 3i)(2 + 3i) = 2 2 (3i) 2 = 3 = z 2 Division of Complex number To divide a complex number by a complex number we use the idea of complex conjugate in order to make the denominator real by rationalizing the denominator. For example, to divide by a complex number a + bi if a + bi 0 we will multiply both numerator and the denominator by the conjugate of a + bi; that is, a + bi bi = ( )(a a + bi a bi ) = a bi a 2 + b 2 a = a 2 + b 2 i b a 2 + b 2
6 Complex Function Example 6.7 Write 2 5i in the form a + ib, where a and b are real numbers. Solution: Multiply the numerator and denominator by the conjugate of 2 5i 2 5i + 5i = ( )(2 2 5i 2 + 5i ) = 2 + 5i 2 2 + 5 2 = 2 29 + 5 29 i Example 6.8 Write 5+4i i in the form a + ib, where a and b are real numbers. Solution: Multiply the numerator and denominator by the conjugate of i 5 + 4i i = ( 5 + 4i )( i i i ) 5i 4i2 = i 2 = 4 5i Example 6.9 Write 7+2i 3 i in the form a + ib, where a and b are real numbers. Solution: Multiply the numerator and denominator by the conjugate of 3 i 7 + 2i 3 i = ( 7 + 2i 3 i )(3 + i 3 + i ) 9 + 3i = 3 2 + 2 = 9 0 + 3 0 i Theorem 6..2 For any three complex numbers z,z 2,andz 3. z Re z z 2. z Im z z 3. z + z 2 z + z 2 Triangular inequality 4. z z 2 z z 2 Polar Form of Complex Number There is a one to one corresponding between the complex number z = x + iy and the order pair (x,y) in the complex plane. The plane on which the complex number are representing is called the complex plane or rectangular form of complex plane. A complex number z = x + iy may also identified as a vector. Figure 6.6: Polar form of complex number
6. Complex Number 7 De-Moiver s Formula Since z and z 2 are complex number z z 2 = (r (cosθ + isinθ )r 2 (cosθ 2 + isinθ 2 ) = r r 2 (cosθ + isinθ )(cosθ 2 + isinθ 2 ) = r r 2 (cos(θ + θ 2 ) + isin(θ + θ 2 )) An induction argument shows that if z i has modules r i and argument θ i, where i =,2,,n, then z z 2 z n = r r 2 r n (cos(θ + θ 2 + + θ n ) + isin(θ + θ 2 + + θ n )) If we let z = z 2 = = z n = z in the above equation, we obtain z n = r n (cosnθ + isinnθ) For z = (the unit circle), then we have (cosθ + isinθ) n = (cosnθ + isinnθ) is called De-Moiver s Formula. Theorem 6..3 De-Moiver s Formula for any n N (cosθ + isinθ) n = (cosnθ + isinnθ) Example 6.0 Simplify ( 3 + i) 7 in rectangular form. Solution: First we have to write z = 3 + i in polar form as z = 2cos π 6 + isin π 6 hence since r = 2 and θ = pi 6, ( 3 + i) 7 = (2e i( π 6 )7 since r=2 and θ = π 6 = 2 7 e i( 7π 6 ) = 2 7 (cos( 7π 6 ) + isin(7π 6 )) = 2 7 (cos(π + π 6 ) + isin(π + π 6 )) = 2 7 ( cos( π 6 ) isin(π 6 )) 3 = 2 7 ( 2 i 2 = 2 6 ( 3 i) = 64 3 64i Example 6. Simplify ( + i) 20 in rectangular form. Solution: First we have to write z = + i in polar form as z = 2cos π 4 + isin π 4 since r = 2 and θ = π 4, hence ( 3 + i) 7 = ( 2e i( π 4 )20 since r=2 and θ = π 6 = 2 20 i( 20π e 4 ) = 2 0 (cos5π + isin5π) = 024( ) = 024
8 Complex Function Example 6.2 Use the De-Movirs Theorem to express cos4θ and sin4θ in terms of powers of cosθ and sinθ. Solution: We know that (cosθ + sinθ) 4 = cos4θ + sin4θ 6cos2θ sin 2 θ + i(4cos 3 θ sinθ 4cosθ sin 3 θ) from De-Movires theorem so from the above two equation (cosθ + sinθ) 4 = cos4θ + sin4θ cos4θ = cos4θ + sin4θ 6cos2θ sin 2 θ sin4θ = 4cos 3 θ sinθ 4cosθ sin 3 θ Roots We can use De-Movere s formula to find the n th roots of a complex number z = r(cosθ + isinθ) write z = w n ; that is w = n z, where w = R(cosφ + isinφ). Thus r(cosθ + isinθ) = R n (cosnφ + isinnφ) By equating the absolute value on both sides in the above equation R n = r n ; that is, R = n r where the root is real positive and thus is uniquely determined. By equating the argument, again we obtain nφ = θ + 2kπ; that is, φ = θ n + 2kπ n where k is an integer. Consequently, n z for z 0, has the n distinct values where k = 0,,2,,n. w = n z = n r(cos( θ n + 2kπ n ) + isin(θ n + 2kπ n )) Theorem 6..4 Let n be positive integers, then any non-zero complex number has n distinct n th roots. Example 6.3 Find 3 + i. Solution: Let z = + i z = 2(cos π 4 + isin π 4 ) and w = 3 z w = R(cosφ + isinφ). So 3 z = w z = (w) 3 π 2(cos 4 + isin π ) = (R(cosφ + isinφ))3 4 π 2(cos 4 + isin π 4 ) = R3 (cos3φ + isin3φ)) 3 2 = R 3 and 3φ = π + 2kπ where k = 0,,2 4 R = 6 2 and φ = π 2 + 2kπ where k = 0,,2 3 If k = 0, then φ 0 = 2 π and w 0 = 6 2(cos 2 π + isin 2 π ). If k =, then φ = 2 π + 2π 3 = 3π 4 and w = 6 2(cos 3π 4 + isin 3π 4 ). If k = 2, then φ 2 = 2 π + 4π 3 = 7π 2 and w 2 = 6 2(cos 7π 7π 2 + isin 2 ). Therefore, w 0, w and w 2 are the third root of z = + i.
6.2 Elementary Functions 9 6.2 Elementary Functions Exponential Function Exponential function can be defined as exp(z) = e z = e x+iy = e x e iy = e x (cosy + isiny), where z = x + iy Example 6.4 Find exp(z + w), where z = x + iy and w = u + iv. Solution: exp(z + w) = e z e w = e x+iy e u+iv = e x (cosy + ie x siny)e u (cosv + ie u sinv) = e x e u (cosy + ie x siny)(cosv + ie u sinv) = e x+u (cos(y + v) + isin(y + v)) Example 6.5 Show that exp(z + i2π) = exp(z). Solution: exp(z + i2π) = e z e i2π = e x+iy e u+iv = e x (cosy + ie x siny)(cos(2π) + isin(2π)) = e x (cosy + ie x siny)( + i0) = e x (cosy + isiny) = e z Trigonometric Function Trigonometric function defined cosine and sine as cosz = eiz + e iz 2 and sinz = eiz e iz 2i where we are using e z = exp(z). Example 6.6 Show that sin(z + 2π) = sinz for all z. Solution: sin(z + 2π) = ei(z+2π) e i(z+2π) 2i = eiz+i2π e iz i2π 2i = eiz e i2π e iz e i2π 2i since e i2π = and e i2π = = eiz e iz 2i = sinz
0 Complex Function Example 6.7 Show that cos(z + 2π) = cosz for all z. Solution: cos(z + 2π) = ei(z+2π) + e i(z+2π) 2 = eiz+i2π + e iz i2π 2 = eiz e i2π + e iz e i2π 2 since e i2π = and e i2π = = eiz + e iz 2 = cosz Example 6.8 Show that sin(z + π 2 ) = cosz for all z. Solution: sin(z + π 2 ) = ei(z+ π 2 ) e i(z+ π 2 ) 2i = eiz+i π 2 e iz i π 2 2i = eiz e i π 2 e iz e i π 2 2i since e i π 2 = i and e i π 2 = i = eiz i e iz ( i) 2i = eiz + e iz 2 = cosz Hyperbolic Function Since for real t Thus, Similarly, cos(it) = cosht. sinht = et e t and cosht = et + e t 2 2 sin(it) = ei(it) e i(it) 2i = i( et e t ) = isinht 2 Logarithmic Function Logarithmic function can be defined as For any value of logz we have. e logz = e ln z +i arg z = e ln z e i arg z = z 2. logz = ln z + iarg z, if z 0 log(e z ) = lne x + i arg i = x + (y + 2kπ)i = z + i2kπ where k is an integer
6.3 Function of Complex Variables, Limit and Derivatives 3. log(zw) = ln( z w ) + iarg(zw) = ln z + iargz + ln w + iargw = logz + logw + 2kπi,for some integer k 6.3 Function of Complex Variables, Limit and Derivatives Corresponding to each value of a complex variable z, there corresponds one value or more than one value of another complex variables w, then w is said to be a function of z and written as w = f (z) = u + iv = u(x,y) + iv(x,y), where x and y are real If to each value of z, there corresponds one and only one value of w, then w is said to be a single valued function of z, otherwise a multiple valued function of z. Example 6.9 w = z,where z 0 is a particular examle of single valued function of z and defined at all points of z-plane except at z = 0. Example 6.20 w = z, where z 0 is a particular examle of multiple valued function of z. Definition 6.3. If a 0,a,a 2,,a n are complex constants, then the function P(z) = a 0 + a z + a 2 z 2 + + a n z n is defined in the entire complex plane and is called a polynomial in z. Definition 6.3.2 If P(x) and Q(x) are two polynomials, then the quotient P(z) Q(z) function and is defined for all z with Q(z) 0. Example 6.2 Find the real and imaginary part of the function f (z) = z 2. Solution: Let w = f (z) = u + vi and z = x + yi, then w = f (z) u + vi = (x + yi) 2 = (x 2 y 2 ) + (2xy)i u(x,y) = x 2 y 2 and v(x,y) = 2xy is called rational Example 6.22 Write f (z) = z 4 in the form of f (z) = u + iv for z = x + iy. Solution: Let z = x + yi, then f (z) = z 4 = (x + iy) 4 = x 4 + 4x 3 yi + 6x 2 (yi) 2 + 4x(yi) 3 + (yi) 4 = (x 4 6x 2 y 2 ) + (4x 3 y 4xy 3 )i Example 6.23 Express f (z) = zre(z) + z 2 + Im(z) in the form of f (z) = u + iv for z = x + iy. Solution: f (z) = zre(z) + z 2 + Im(z) = (x iy)x + (x 2 y 2 + 2xyi) + y = (2x 2 y 2 + y) + (xy)i
2 Complex Function Example 6.24 Express f (z) = 4x 2 + i4y 2 by formula involving the variable z and z. Solution: Let z = x + yi. Then z = x yi. This implies that x = z + z 2 and y = z z 2i Therefore, we have f (z) = 4( z + z 2 )2 + 4( z z ) 2 i 2i = z 2 + 2z z + z 2 (z 2 + 2z z + z 2 )i = ( i)z 2 + (2 + 2i)z z + ( i) z 2 Limit Limit of complex functions have the same properties as that of vector valued function since complex function also considered as a vector valued function. Definition 6.3.3 Let S be an open subset of the complex plane C. Let f be a function on S and L be a complex constant we say that lim z z 0 f (z) = L if for every ε > 0, there is δ > 0 such that if z S and 0 < z z 0 < δ, then f (z) L < ε. Geometrically the definition states that given any open disk with center L and radius ε, there exist an open disk with center z 0 and radius δ such that for all points of z( z 0 ) in the disk 0 < z z 0 < δ, the image w = f (z) lies in the disk f (z) L < ε. Figure 6.7: The image of f (z) Remark 6.3. If a function f (z) approaches to two different values as z z 0 along two different paths, then lim z z0 f (z) does not exist. Example 6.25 Show that lim z 0 z z does not exist.
6.3 Function of Complex Variables, Limit and Derivatives 3 Solution: If z 0 along the line y = mx, where m R and z = x + yi, we have z lim z 0 z (x,y) (0,0) x yi x + yi x (mx)i since y = mx x 0 x + (mx)i x( mi) x 0 x( + mi) mi x 0 + mi = mi + mi If the value of m varies, the limit also varies. Therefore, lim z 0 z z does not exist. Example 6.26 Find the limit of f (z) = (x+y)2 x 2 +y 2 as z 0. Solution: Now we have to show the limit of the function as z 0 from any direction in the complex plane C. So But along the path y = mx, we have lim (lim f (z)) = x 0 y 0 lim (lim f (z)) = y 0 x 0 (x + mx) 2 lim f (z) z 0 x 0 x 2 + (mx) 2 = ( + m)2 + m 2 The limiting value here depends on m and hence lim z 0 f (z) does not exist. Theorem 6.3.2 Let f (z) = u(x,y) + v(x,y)i be a complex function that is defined in some neighborhood of z 0, except perhaps z 0 = x 0 + y 0 i. Then lim f (z) = w 0 = u 0 + v 0 i z z 0 if and only if lim u(x,y) = u 0 and lim v(x,y) = v 0 (x,y) (x 0,y 0 ) (x,y) (x 0,y 0 ) Theorem 6.3.3 Let f and g be two functions whose limits at z 0 exists such that lim f (z) = w 0 and z z0 lim g(z) = k 0, then z z 0. lim [ f (z) ± g(z)] = w 0 ± k 0 z z0 2. lim f (z) g(z) = w 0 k 0 z z0 f (z) 3. lim z z0 g(z) = w 0 k 0, where k 0 0 Theorem 6.3.4 Let f (z) be a complex function. lim f (z) = lim z z0 z z0 f (z) = 0 2. lim f (z) = w 0 lim f ( z z 0 z ) = w 0
4 Complex Function 3. lim f (z) = w 0 lim = 0 z z 0 f ( z ) Example 6.27 Evaluate lim Solution: Since lim z+ z iz+ z iz+ z+. = 0, then lim z iz+ z+ =. Theorem 6.3.5. If lim f (z) = L, then lim f (z) = L z z0 z z0 2. If lim f (z) = L, then lim f (z) = L z z0 z z0 3. lim f (z) = L if and only if lim Re f (z) = Re L and lim Im f (z) = Im L z z0 z z0 z z0 Example 6.28 Evaluate lim z 2i (2x + iy) 2. Solution: lim (2x + z 2i iy)2 4x 2 + i4xy y 2 z 2i (x,y) (0,2) 4x2 y 2 + i lim 4xy (x,y) (0,2) = 4 Continuity Definition 6.3.4 Let f be a complex valued function defined in a region S of the complex plane. Then f is said to be continuous at z 0 if lim f (z) = f (z 0 ). Thus f is continuous at z 0 if given ε > 0 z z0 there exist a δ > 0 such that z z 0 < δ f (z) f (z 0 ) < ε Definition 6.3.5 A function f is continuous at a point z 0, if the following three conditions are satisfied. lim z z0 f (z) exists 2. f (z 0 ) exists 3. lim z z0 f (z) = f (z 0 ) A complex function f is continuous if and only its real and imaginary parts u and v are continuous. Continuity of complex functions has the same properties as that of real functions. Theorem 6.3.6 If f and g are continuous at z 0, then the following functions are continuous at z 0. Their sum and difference f (z) ± g(z) 2. Their product f (z) g(t) 3. Their quotient f (z) g(z), provided that g(z 0) 0 4. Their composition f (g(z)), provided that f (z) is continuous in a neighborhood of the point g(z 0 ). Example 6.29 Show that lim Solution: z +i z 2 2i z 2 2z+2 = i. lim z +i z 2 2i z 2 2z + 2 (z i)(z + + i) z +i (z i)(z + i) z + + i z +i z + i = i
6.3 Function of Complex Variables, Limit and Derivatives 5 Example 6.30 Show that lim z 0 f (z) does not exist, where Solution: Along the line y = x Along the line y = 2x f (z) = z2 z 2 z 2 lim f (z) z 0 z 0 z 2 x 2 y 2 + 2xyi (x,y) (0,0) x 2 + y 2 2x 2 i (x,y) (0,0) 2x 2 = i z 2 lim f (z) z 0 z 0 z 2 x 2 y 2 + 2xyi (x,y) (0,0) x 2 + y 2 x 2 + i4x 2 (x,y) (0,0) 5x 2 = 5 + i4 5 since the two limits are different, we conclude that the limit of f (z) at z = 0 does not exist. Theorem 6.3.7 If f (z) is continuous in a closed and bounded region R, then f (z) reaches a maximum value some where in R. Derivatives Definition 6.3.6 Let w = f (z) be a single value function defined in a domain S and let z 0 be any fixed point in S. The derivatives of f at z 0 written as f (z 0 ) is defined by provided this limit exists. f (z 0 ) z z0 f (z) f (z 0 ) z z 0 Example 6.3 If f (z) = z 3, then show that f (z) = 3z 2. Solution: f (z 0 ) z z0 z 3 z 3 0 ) z z 0 z z0 (z z 0 )(z 2 + zz 0 + z 2 0 ) (z z 0 ) z z0 z 2 + zz 0 + z 2 0 = 3z 2 0 since z 0 is arbitrary, it follows that f (z) = 3z 2.
6 Complex Function Example 6.32 Show that the function w = f (z) = z = x yi is no where differentiable. Solution: Along the line parallel to the x-axis Along the line parallel to the y-axis z z 0 lim z z 0 z z 0 (x,y) (x 0,y 0 ) = (x,y) (x 0,y 0 ) z z 0 lim z z 0 z z 0 (x,y) (x 0,y 0 ) (x yi) (x 0 + y 0 i) (x + yi) (x 0 + y 0 i) x x 0 x x 0 (x yi) (x 0 + y 0 i) (x + yi) (x 0 + y 0 i) (y y 0 ) (x,y) (x 0,y 0 ) i(y y 0 ) = since z 0 is arbitrary, it follows that it is no where differentiable. Theorem 6.3.8 If f is differentiable at z 0, then f is continuous at z 0. Basic Properties of Derivatives. d dz C = 0, where C is a constant. 2. d dz z n = nz n, where n is a positive integral. 3. d dz C f (z) = C f (z) 4. d dz [ f (z) ± g(z)] = f (z) ± g (z) 5. d dz ( f (z) g(z)) = f (z)g(z) + f (z)g (z) 6. d dz ( f (z) g(z) ) = f (z)g(z) f (z)g (z), provided that g(z) 0 (g(z)) 2 7. d dz f (g(z)) = f (g(z))g (z) Example 6.33 Find d dz (z2 + 2zi + 3) 4. Solution: d dz (z2 + 2zi + 3) 4 = 4(z 2 + 2zi + 3) 3 d dz (z2 + 2zi + 3) = 4(z 2 + 2zi + 3) 3 (2z + 2i) = 8(z 2 + 2zi + 3) 3 (z + i) If n is a positive integer, then f (z) = z n is differentiable in the entire complex plane whereas f (z) = Rez, f (z) = Imz and f (z) = z are not differentiable but f (z) = z is differentiable in C {0} on the other hand f (z) = z 2 is differentiable only at z = 0. 6.4 Analytic Functions and Their Derivatives A function f is said to be analytic at a point z 0, if it is differentiable throughout some ε-neighborhood of z 0. A function f is analytic in a region if it is analytic at every point of the region. A function which is differentiable at a point, need not necessarily be analytic at that point, for example f (z) = z 2 is differentiable only at z = 0. Definition 6.4. Let f (z) = u(x,y) + iv(x,y), where the partial derivatives of u and v with respect to
6.4 Analytic Functions and Their Derivatives 7 x and y exist and denoted as u x, v x, u y and v y. Then the Cauchy-Riemann equation is satisfied as u x = v y and u y = v x Theorem 6.4. f (z) = u(x,y) + iv(x,y) is analytic or differentiable at a point z 0, if the partial derivatives of u and v exists, are continuous and satisfies the Cauchy-Riemann condition. Theorem 6.4.2 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 exist at (x 0,y 0 ), and they must satisfy the Cauchy-Riemann equation there also, f (z 0 ) can be written u x = v y and u y = v x f (z 0 ) = u x + iv x where these partial derivatives are to be evaluated at (x 0,y 0 ). Example 6.34 Show that f (z) = z 2 is analytic. Solution: Let z = x + yi, f (z) = z 2 = (x 2 y 2 ) + (2xy)i u(x,y) = x 2 y 2 and v(x,y) = 2xy u x (x,y) = 2x and u y (x,y) = 2y v x (x,y) = 2y and v y (x,y) = 2x Therefore, u x = v y and u y = v x. Here Cauchy-Riemann equation is satisfied and u x, v x, u y, v y are exists and also the partial derivatives are continuous. Hence f (z) = z 2 is analytic or differentiable. Example 6.35 Show that f (z) = z 3 is analytic. Solution: Let z = x + yi, f (z) = z 3 = (x 3 3xy 2 ) + (3x 2 y y 3 )i u(x,y) = x 3 3xy 2 and v(x,y) = 3x 2 y 3 u x (x,y) = 3x 2 3y 2 and u y (x,y) = 6xy v x (x,y) = 6xy and v y (x,y) = 3x 2 3y 2 Therefore, u x = v y and u y = v x. Here Cauchy-Riemann equation is satisfied and u x, v x, u y, v y are exists and also the partial derivatives are continuous. Hence f (z) = z 3 is analytic or differentiable. Example 6.36 Show that f (z) = z 2 is analytic. Solution: Let z = x + yi, f (z) = z 2 = x 2 + y 2 u(x,y) = x 2 + y 2 and v(x,y) = 0 u x (x,y) = 2x and u y (x,y) = 2y v x (x,y) = 0 and v y (x,y) = 0 Here Cauchy-Riemann equation is not satisfied. Therefore, f (z) = z 2 is not analytic.
8 Complex Function Definition 6.4.2 f (x,y) is a harmonic function if and only f x, f y, f xx and f yy exist and continuous and satisfies the condition f xx + f yy = 0 This equation is known as Laplace equation. Definition 6.4.3 If f (x, y) = u(x, y) +iv(x, y), satisfies Laplace equation, then u is harmonic conjugate of v and v is also harmonic conjugate of u. Example 6.37 Show that f (z) = e z is analytic and f (z) = e z. Solution: Let z = x + yi, f (z) = e z = e x+yi = e x e iy = e x (cosy + isiny) = e x cosy + ie x siny u(x,y) = e x cosy and v(x,y) = e x siny u x (x,y) = e x cosy u xx (x,y) = e x cosy u y (x,y) = e x siny u yy (x,y) = e x cosy v x (x,y) = e x siny v xx (x,y) = e x siny v y (x,y) = e x cosy v yy (x,y) = e x siny Here, u x, v x, u y, v y, u xx, v yy are exists and also the partial derivatives are continuous and u xx + u yy = 0 and v xx + v yy = 0 Therefore u and v are harmonic. And Cauchy-Riemann equation is satisfied since u x = v y and u y = v x. Therefore v is a harmonic conjugate of u. Hence f (z) = z 2 is analytic or differentiable. In addition to this f (z) = u x + iv x = e x cosy + ie x siny = e z Example 6.38 Show that f (z) = e z is not analytic. Solution: Let z = x + yi, then z = x yi f (z) = e z = e x yi = e x cosy ie x siny u(x,y) = e x cosy and v(x,y) = 3x 2 y 3 u x (x,y) = e x cosy and u y (x,y) = e x siny v x (x,y) = e x siny and v y (x,y) = e x cosy Here, u x v y and u y v x. So, Cauchy-Riemann equation is not satisfied.therefore f (z) = z 3 is not analytic. Example 6.39 Show that e z = e z. Solution: Let z = x + yi, then z = x yi e z = e x yi = e x cosy ie x siny () e z = e x cosy + ie x siny = e x cosy ie x siny (2) e z = e z from equation () and (2)
6.5 Line Integral in the Complex Plane 9 6.5 Line Integral in the Complex Plane Definition 6.5. Let S C be a region. Let [a,b] be some interval. A path (curve or contour) in S is defined to be a continuous function : [a,b] S we call (a) the initial point and (b) the terminal point. Definition 6.5.2 (t) is said to be a smooth curve if (t) is continuous and (t) 0, since (t) is continuous in [a,b], then the smooth arc has length L; that is, where L is arc length. b L = (t) dt a Definition 6.5.3 A path is said to be rectifiable if L() <. Definition 6.5.4 If is piecewise differentiable, then is rectifiable and L() = b a (t) dt. Definition 6.5.5 Let : [a,b] C be a rectifiable path with S C and f : S C be defined and continuous on the trace. Then the line-integral of f along is defined by the equation denoted this integral by b a b f ((t))d(t) = ( f o)d a f (z)dz = f dz Definition 6.5.6 A curve is a piecewise smooth curve, that is it consists of a finite number of smooth arcs joined end to end. Figure 6.8: Piecewise smooth curve Example 6.40 Consider z(t) such that z(t) = joined end to end. { t + it t [0,] t + i t [,2] is piecewise smooth curve and Definition 6.5.7 Let C be a piecewise differentiable curve given by the equation z = z(t),wherea t b. Let f (z) be a continuous complex valued function defined in a region containing the curve C, we define b f (z)dz = f (z(t))z (t)dt C a Example 6.4 Let : [0,2π] C be given by (θ) = e iθ and defined f (z) = z the integral of f over the circle. for z 0. Evaluate
20 Complex Function Solution: We have to evaluate the integral of f over the circle; that is, by definition, this integral is equal to 2π 0 dz where 0 θ 2π z 2π e iθ ieiθ dθ = i dθ = i2π 0 Remark 6.5.. We call a differentiable path if (u) exist for each u [a,b] and : [a,b] C is continuous. 2. is said to be piecewise differentiable if there is a partition on [a,b] such that is differentiable on each sub-interval. 3. The set { : a t b} is called the trace of where : [a,b] C is a path. 4. The trace is denoted by {} and the length is denoted by L(). Definition 6.5.8 The curve C is said to be simple if t t 2, then (t ) (t 2 ); that is, it does not cross it self. Definition 6.5.9 The curve C is called a closed curve if (a) = (b) but the curve C is called a simple closed curve if (a) = (b) and (t ) (t 2 ) for any other distinct real numbers t,t 2 [a,b] and t t 2. Figure 6.9: Simple and non simple curves Remark 6.5.2 A simple closed curve is also called a Jordan Curve. Example 6.42 Show that the curve (t) = cost + isint,0 t 2π is simple closed curve. Solution: since (t ) (t 2 ) for any other distinct real numbers t,t 2 (0,2π) and (0) = (2π) =. Then the circle is simple closed curve. Properties of Line Integral. ( f (z) + g(z))dz = f (z)dz + g(z)dz. It can be easily generated for a finite number of functions. 2. f (z)dz = f (z)dz, where traversed in the opposite direction of 3. f (z)dz = f (z)dz + f (z)dz, where the final point of coincides with the initial point of + 2 2 2. In general, if = + 2 + + n, where the final point of k coincides with the initial point of k+,k =,2,,n, then f (z)dz = f (z)dz + f (z)dz + + 2 n f (z)dz
6.6 Cauchy s Integral Theorem and Formula 2 4. C f (z)dz = C f (z)dz, where C is any complex constant. 5. (C f (z)dz +C 2 f 2 (z)dz + +C n f n (z)dz) = C f (z)dz +C 2 6. An improper inequality f (z)dz f (z) dz f 2 (z)dz + +C n f n (z)dz 6.6 Cauchy s Integral Theorem and Formula Definition 6.6. A simple connected region R is a region such that every simple closed contour with in it encloses only the point of R; that is, R = int of Definition 6.6.2 If a region is not simply connected, then it is multiply connected; that is, R int of Figure 6.0: Simple and non simple connected region Theorem 6.6. If a function is analytic everywhere, throughout a simply connected domain, then any simple closed contour in R f (z)dz = 0 Figure 6.: Simple connected region D Theorem 6.6.2 Green s Theorem For real valued Function If is a closed contour and P(x,y) and Q(x,y) have continuous partial derivatives, then where R is called interior to and on. (Pdx + Qdy) = (Q x P y )dxdy R
22 Complex Function Applying green s theorem for evaluating complex functions from contour integral b f (z)dz = f ((t)) (t)dt = 0 where a t b a Theorem 6.6.3 If a function f is analytic at a point z 0, then its derivatives of all orders exist and are also analytic functions at that point and f (n) z 0 = n! 2πi f (z) dz, where n z+ (z z 0 ) n+ Example 6.43 Find z π dz, where z = in C. z 3 Solution: f (z) = z π is analytic on C and interior of C. Here n + = 3 n = 2. z π z π z 3 dz = (z 0) 3 dz = 2πi 2! f (2) (0) = 2πi 2! (0) = 0 Example 6.44 Find e 2z dz, where z = 4 in C. z 4 Solution: f (z) = e 2z is analytic on C and interior of C. Here n + = 4 n = 3. e 2z z 4 dz = e 2z (z 0) 4 dz = 2πi 3! f (3) (0) = 2πi 6 (8e0 ) = 8π 3 i Theorem 6.6.4 Morero s Theorem If a function is continuous throughout a region R and if f (z)dz = 0 for every closed contour on R, then f is analytic throughout R. Theorem 6.6.5 Cauchy-Goursat Theorem Let T be a rectangle and let f be analytic on T, then δ f = 0 Example 6.45 (3z+)dz, where is the sides of a square with vertices (0,0),(,0),(0,) and (,). Solution: (3z + )dz = (3x + )dx + (3( + iy) + )dy + = 0 0 0 (3(x + i) + )dx + 0 0 (3yi + )dy
6.7 Sequence and Series 23 is a simple closed contour and f (z) = 3z + is analytic everywhere. Hence, using Cauchy-Goursat s theorem (3z + )dz = 0 Theorem 6.6.6 Cauchy s Integral Formula Let f be analytic everywhere inside and on a simple closed contour, taken in the positive sense. If z 0 is any point interior to, then f (z 0 ) = 2πi f (z) z z 0 dz Example 6.46 Let C be the positive oriented circle z = 2, since the function f (z) = z 9 z 2 is analytic within and on C and since the point z 0 = i is interior to. So, z z (9 z 2 )(z + i) C dz = 9 z 2 dz = 2πi( i z ( i) 0 ) = π 5 6.7 Sequence and Series Definition 6.7. An infinite sequence z,z 2,,z n, of a complex numbers has a limit z if for each positive ε, there exist a positive integer n 0 such that z z n < ε whenever n > n 0 In other words, for large value of n the points z n are arbitrary close to the point z. Theorem 6.7. The sequence can have at most one limit. That is, a limit z is unique if it exists when that limit exists,the sequence is said to converge to z and we write If the sequence has no limit, it diverges. lim z n = z n Theorem 6.7.2 Suppose that z n = x n + iy n, where n =,2,3 and z = x + iy. Then lim z n = z if and only lim x n = x and lim y n = y n n n Definition 6.7.2 An infinity series z + z 2 + z 3 + + z n + of complex numbers converges to a sum S if the sequence s n = n i= z i of partial sums converges to s. we then write n= z n = s. Remark 6.7.3. As a sequence can have at most one limits a series can have at most one sum. If the series does not converges it is said to be divergent. n 2. A series z n converges to s if for every ε > 0, there is a positive integer n 0 such that z i i=0 s < ε, whenever n > n 0. If this is the case, we say that s is the sum of the infinite series and denoted by s = z i. i=0
24 Complex Function 3. A necessary condition for the converges series i=0 z i is that lim n z n = 0. Theorem 6.7.4 The terms of a convergent series of complex numbers are bounded. Definition 6.7.3 The series z n converges absolutely if z n converges. Test of convergence These tests are applicable to series whose terms are either complex or real( or even for functions of complex variables) Theorem 6.7.5. Ratio test Let f n (z) = f (z) + f 2 (z) + + f n (z) + be a series of non zero functions of complex variables and lim f n+ (z) n f n (z) = Φ(z) Then (a) The series converges absolutely for these value of z for which 0 < Φ(z) <. (b) The series diverges for these value of z for which Φ(z) >. (c) No information can be drawn from the test for those of z for which Φ(z) =. 2. Cauchy s root test Let variables and f n (z) = f (z) + f 2 (z) + + f n (z) + be a series of non zero functions of complex lim n f n+ (z) f n (z) n = Φ(z) Then (a) The series converges absolutely for these value of z for which Φ(z) <. (b) The series diverges for these value of z for which Φ(z) >. (c) No information can be drawn from the test for those of z for which Φ(z) =. Definition 6.7.4 A representation of an analytic function by infinite series of complex function are called Taylor series. Every analytic function can be represented by a power series and are quiet similar to the familiar series of real valued function. Definition 6.7.5 A power series about z 0 is an infinite series of the form a n (z z 0 ) n = a 0 + a (z z 0 ) + a 2 (z z 0 ) 2 + Where z 0 and a 0,a,a 2, and complex numbers. If z 0 = 0, in the above series we obtain as a particular case of a power series Suppose the power series a n z n = a 0 + a z + a 2 z 2 + a n (z z n ) n = a 0 + a (z z 0 ) + a 2 (z z 0 ) 2 +, we would like to consider all the points z in the complex plane for which the series converges. The following three possibility arise. The series converges only at z = 0 2. The series converges in the entire complex plane.
6.7 Sequence and Series 25 3. There exists an open disk z z 0 = R such that the series converges in side the disk and diverges out side the disk. Its radius R is called the radius of convergence for the series a n (z z n ) n. Figure 6.2: Radius of convergency Example 6.47 Identify the region where z n n! Solution: Using ratio test for convergence, we obtain since 0 <, then the series the series series z n z n z n n! converges and diverges. lim z n+ n! n (n + )! z n z n n + z n n + = 0 converges absolutely for all z in the entire complex plane. Consequently, converges in the entire complex plane which means the radius of convergence for the n! is R =. n! ( ) n z 2n+ Example 6.48 Identify the region where converges and diverges. (2n + )! Solution: Using ratio test for convergence, we obtain since 0 <, then the series Consequently, the series ( ) n+ z 2(n+)+ lim n (2(n + ) + )! convergence for the series ( ) n z 2n+ (2n + )! ( ) n z 2n+ (2n + )! ( ) n z 2n+ (2n + )! Example 6.49 Identify the region where (2n + )! ( ) n z (2n+) z 2 n (2n + 3)(2n + 2) n z 2 (2n + 3)(2n + 2) = 0 converges absolutely for all z in the entire complex plane. converges in the entire complex plane which means the radius of is R =. n!z n converges and diverges.
26 Complex Function Solution: Using ratio test for convergence, we obtain lim (n + )!z n+ n n!z n (n + )z n Therefore, z n + n { if z 0 = 0 if z = 0 n!z n converges at z = 0 and hence its radius of convergence is R = 0. z n Example 6.50 Identify the region where converges and diverges. n Solution: Using ratio test for convergence, we obtain lim z n+ n n (n + ) z n z(n) n n + z n n n + = z lim n n n + = z z n Therefore, the series converges for z < and the series diverges z >. Furthermore, at z = the n series diverges and at z = the series converges. Theorem 6.7.6 Suppose that a function f is analytic throughout a disk z z 0 < R 0, centered at z 0 and with radius R 0. Then f (z) has the power series representation f (z) = where a n = f (n) (z 0 ) n! with n = 0,,2,. That is, series the stated open disk. Notice that f 0 (z 0 ) = f (z 0 ) and 0! = so series f (z) = f (z 0 ) + f () (z 0 )! a n (z z 0 ) n,( z z 0 < R 0 ) a n (z z 0 ) n converges to f (z) when z lies in a n (z z 0 ) n can be written as + f (2) (z 0 ) 2! +,( z z 0 < R 0 ) Any function which is analytical at a point z 0 must have a Taylor series about z 0. If z 0 = 0, in which case f (n) (0) series becomes f (z) = (z) n,( z < R 0 ) is called a Maclaurin Series. n! Theorem 6.7.7 If f is entirly analytic and bounded in the complex plane, then f (z) is constant throughout the plane. Theorem 6.7.8 Suppose that a function f is analytic inside and on a positive oriented circle C R, centered at z 0 and with radius R. If M R denoted the maximum value of f (z) on C R, then f (n) (z 0 ) n!m R R n where n =,2,3, ()
6.8 Laurent s Theorem 27 Inequality () is called Cauchy s Inequality. Theorem 6.7.9 Any polynomial P(z) = a 0 + a z + + a n z n (a n 0) of degree n(n ) has at least one zero. That is, there exist at least one point z 0 such that P(z 0 ) = 0. 6.8 Laurent s Theorem If a function f fails to be analytic at a point z 0, we cannot apply Taylor s theorem at that point. It is often possible, however, to find a series representation for f (z) involving both positive and negative powers of z z 0 ; that is, f (z) = a n (z z 0 ) n + b n (z z 0 ) n Theorem 6.8. Laurent s series expansion theorem Let C and C 2 be two concentric circles having center at z 0 and radius r and r 2 respectively with r < r 2. Suppose f be single valued function on the circles and with in the annuls between C and C 2, then f (z) = C n (z z 0 ) n where C n = 2πi f (t) δ dt with δ is any circle lying between C (t z 0 ) n+ and C 2 having z 0 as a center for all values of n or f (z) = a n (z z 0 ) n + n= b n (z z 0 ) n Figure 6.3: Concentric circle Example 6.5 Find the Laurent s series for z(z ) 2 at the point z =.
28 Complex Function Solution: Putting z = z, then z(z ) 2 = (z + z )(z ) 2 = z 2 ( + z ) = z 2 [ z + z 2 z 3 + ] using binomial expansion = z 2 z + z + z 2 = (z ) 2 (z ) + (z ) + (z )2 Thus, z(z ) 2 = ( )n (z ) n 2. This expansion is valued in the region out side the circle with center at the point and z = 0 should not be a point with in the circle z <. That is, z(z ) 2 = ( ) n (z ) n 2, z < 6.9 Singularity and Zeros, Infinity Definition 6.9. If a function f fails to be analytic at a point z 0 but is analytic at some point in every neighborhood of z 0, then z 0 is called a singular point, or singularity of f. The point z = 0 is a singular point of a function f (z) = z.the function f (z) = z 2, on the other hand, has no singular points since it is nowhere analytic. Definition 6.9.2 A singular point z 0 is said to be isolated if, in addition, there is a deleted neighborhood 0 < z z 0 < ε of z 0 throughout which f is analytic. Example 6.52 The function f (z) = z+ has the three isolated singular points z = 0 and z = ±i. z 3 (z 2 +) Let D is the domain between two concentric circles at a point a b n f (z) = a n (z z 0 ) n + = R(z z 0 ) + P( ) z z 0 b n (z z 0 ) n where P( z z 0 ) = (z z 0 ) n is called the principal part and R(z z 0) = part of f (z) in the neighborhood of z 0. a n (z z 0 ) n is the regular Definition 6.9.3 If f (z) is analytic in the domain D except at z = z 0 ; that is, f (z) is analytic in the deleted neighborhood of z 0, then the above Laurent s expansion is valid in such a domain and. If z 0 is an isolated singularity of f and if all but a finite number of the b n are zero, then z 0 is called a pole of f. If k is the highest integer such that b k 0, z 0 is called a pole of order k. In particular, if z 0 is a first order pole, then it is called a simple pole. 2. If an infinite number of b ns are nonzero, then z 0 is called an essential singularity. 3. The coefficient b of z a is called the residue of f at z 0. 4. If all the b ns are zero, we say that z 0 removable singularity.
6.0 Residue Integration Method 29 6.0 Residue Integration Method If f has isolated singularity at z 0, then f admits a Laurent expansion that is valid in a deleted neighborhood of z 0 f (z) = + b 2 (z z 0 ) 2 + b + a 0 + a (z z 0 ) + z z 0 Where b is called the residue of f at z 0. This is written as b = Res( f,z 0 ). Now we want to develop techniques for computing the residue with out having to find the whole Laurent expansion. Of course, if the Laurent expansion is known, there is no problem. Example 6.53 Since e z = + z + + + 2z 2 n!z n + the coefficient of z is, So, f (z) = e z has residue at z 0 = 0 Residue values If the residue of f at z 0 will be the coefficient of z z 0 in the expansion. If not, consider the following cases Case : Let the order of z 0 be one; that is, the pole is simple pole, then Therefore, f (z) = whenever z = z 0 is simple pole. Case 2: Let the order of z 0 be m, then Therefore, f (z) = Res( f (z),z 0 ) = b = a n (z z 0 ) n + b z z 0 Res( f (z),z 0 ) = b z z0 (z z 0 ) f (z) a n (z z 0 ) n + b z z 0 + b 2 (z z 0 ) 2 + + b m (z z 0 ) m (m )! lim z z 0 (z z 0 ) dm dz m (z z 0) m f (z)) whenever z = z 0 is a pole of order m. Let g and h be analytic at z 0 and assume that g(z 0 ) 0, h(z 0 ) = 0 and h (z 0 ) 0, then f (z) = g(z) h(z) has a simple pole at z 0 and Res( f,z 0 ) = g(z 0) h (z 0 ) Remark 6.0.. If the function f(z) has no poles, then the residue equal to zero. 2. If a function f fails to be analytic at a point z 0 but is analytic at some point in every neighborhood of z 0, then z 0 is called a singular point, or singularity of f. 3. A singular point z 0 is said to be isolated, if there is a deleted neighborhood 0 < z z 0 < ε of z 0 throughout which f is analytic. Theorem 6.0.2 Residue Theorem If C is simple closed curve and if f (z) is analytic within and on C except at a finite number of poles or isolated singularity with C, then C f (z)dz = 2πi(R + R 2 + + R n ) = 2πi where R,R 2,,R n are residues of f (z) at these points in C. n R k k=
30 Complex Function Figure 6.4: Region with a finite number of poles Example 6.54 Evaluate C Solution: f (z) = 2z2 +z z 2 = outside of C : z =. Therefore, Where 2z 2 +z dz,c : z =. z 2 z(2z+) (z )(z+) dz. Hence it has poles at z = and z =. But z = is a pole C: z = R = Res( f,) 2z 2 + z z 2 dz = 2πiR 2z 2 + z (z )( z (z )(z + ) ) z(2z + ) ( z (z + ) ) = 3 2 Hence, C 2z 2 +z z 2 dz = 2πi( 2 3 ) = 3πi where C : z =. Example 6.55 Evaluate C Solution: f (z) = 2z2 +z z 2 = 2z 2 +z dz, C : z = 2. z 2 z(2z+) (z )(z+) dz. Both poles z = and z = lie within the circle C : z = 2. Thus, 2z 2 + z Res( f, ) (z )( z (z )(z + ) ) z(2z + ) ( z (z + ) ) = 3 2 + ) Res( f,) (z + )(z(2z z (z + ) ) = 2 Hence, C 2z 2 +z z 2 dz = 2πi( 2 3 + ( 2 )) = 2πi where C : z = 2. Evaluation of the definite Integrals Integration of the integer 2π 0 F(cosθ,sinθ)dθ, where F(cosθ,sinθ) is a rational function of cosθ and sinθ that is finite in the interval 0 θ 2π.Consider a unit circle C : z =, hence, z = e iθ, where
6.0 Residue Integration Method 3 0 θ 2π. Thus, dz = ie iθ dθ, which gives dθ = dz = dz ie iθ iz. Moreover, z = eiθ = cosθ + isinθ and z = e iθ = cosθ isinθ. Therefore, cosθ = 2 (z + z ) and sinθ = 2i (z z ). Hence, 2π 0 F(cosθ,sinθ)dθ = F( C: z = 2 (z + z ), 2i (z z ))dz iz C = f (z)dz where C is a unit circle with its center at the origin. Hence, using residue s theorem, it is equal to 2πi(R) is the sum of the residue of f (z) within C. Example 6.56 Evaluate 2π 0 dθ +asinθ, where < a <. Solution: Consider a unit circle C : z =. Hence, z = e iθ, where 0 θ 2π. Thus, dz = ie iθ dθ and sinθ = 2i (z z ) = z2 2iz. Hence, Now, consider f (z) = 2π 0 dθ + asinθ dz iz = C + a( 2i ( z2 z )) 2izdz = C iz(2iz + az 2 a) dz = 2 az 2 where C : z = + 2iz a C dz az 2 +2iz a. The poles of f (z) are at az2 + 2iz a = 0 which gives us z = 2i + 2i a 2 2a z 2 = 2i 2i a 2 2a = i( + a 2 ) a = i( a 2 ) a Now, z < for < a < and z 2 > for < a <. Therefore, the only pole lies inside C : z = is z and it is simple pole. Thus, and Therefore, Res( f,z ) z z (z z ) f (z) 2π 0 dθ + asinθ (z z ) z z a(z z )(z z 2 ) z z a(z z 2 ) = a(z z 2 ) = 2i a 2 = 2 a 2 dz = 2 C az 2 + 2iz a = 2(2πi)( 2 a 2 ) 2π =, < a < a 2
32 Complex Function 6. Exercise. Prove that cos5θ = 6cos 5 θ 20cos 3 θ + 5cosθ. 2. Prove that sin5θ sinθ = 6cos4 θ 2cos 2 θ +. If θ o,±π,±2π,. 3. Evaluate ( +i 3 i 3 )0. 4. Find the square root of 5 i8. Ans. 2 + i 3 2 Ans. 5. Solve the equation z 2 + (2i 3)z + 5 i = 0. Ans. z = 2 i3 or z = + i 6. Evaluate lim z 2e iπ 3 z 3 +8 z 2 +4z 2 +6. 7. Show that at z = i, the function f (z) = 3z4 2z 3 +8z 2 2z+5 z i 3z 4 2z 3 + 8z 2 2z + 5 lim z i z i Ans. 3 8 i 3 8 is removable discontinuouity and evaluate Ans. 4 + i4 8. For what value of z are the function f (z) = z z 2 + is continuous. Ans. except z = ±i 9. Prove that + z + z 2 + z 3 + = z if z <. 0. Find the derivatives of ztan (lnz).. Evaluate lim z i z 0 + z 6 +. 2. Evaluate lim(cosz) z 3. z 0 Ans. +(lnz) 2 + tan (lnz) Ans. 5 3 Ans. 2 3. Locate and name all the singularities of f (z) = z8 +z 4 +z (z ) 3 (3z+2) 2 and determinant where f (z) is analytic. Ans. Singularities at z = and z = 2 3 ; z = is a pole of order 3 and z = 3 2 is a pole of order 2 and it is analytic everywhere in the finite z plane except at the points z = and z = 2 3. 4. Verify that the Cauchy-Riemann equation are satisfied for functions f (z) = e z2. 5. Evaluate (2,4) (0,3) (2y + x2 )dx + (3x y)dy along the parabola x = 2t, y = t 2 + 3. 6. Along the straight lines from (0,3) to (2,3) and then from (2,3) to (2,4). Ans. 33 2 Ans. 5 2
6. Exercise 33 7. Evaluate Green s Theorem in the plane for (2xy x 2 )dx + (x + y 2 )dy where C is closed curve of the region bounded by y = x 2 and y 2 = x. C Ans. 30 8. Evaluate C dz (z a) n, n = 2,3,4,, where z = a is inside the simple closed curve C. Ans. 0 9. If C is the curve y = x 3 3x 2 + 4x joining points (,) and (2,3), find the value of C (2z2 4iz)dz. 20. Find the region of convergence of the series n= ( ) n z 2n (2n )!. Ans. 56 + i38 Ans. The series absolutly converges for all z.