Complex Analysis I MAST31006

Transcription

1 Complex Analysis I MAST36 Lecturer: Ritva Hurri-Syrjänen Lectures: Tuesday: :5-2:, Wednesday: 4:5-6: University of Helsinki June 2, 28

2 Complex Analysis I CONTENTS Contents Background 4. The vector space R The complex plane C The elementary properties of complex numbers 8 2. Complex numbers can be written in Cartesian form Theorem 2..: The triangle inequality Complex numbers can be written in polar form Definition 2.2.: Polar coordinates Theorem 2.2.4: Euler s formula The geometric meaning of the multiplication of complex numbers Theorem 2.2.5: De Moivre s theorem Applications of De Moivre s theorem The n th roots of complex numbers Special case: the n th roots of unity About history Topological notions of sets in the complex plane 9 3. Sets in the complex plane Connectedness Limits and continuity Sequences Limits of functions Continuity The Cauchy convergence principle for C Analytic functions Definitions Elementary properties Theorem 4.2.: Chain rule Theorem 4.2.: Derivative of inverse function The relationship between the complex and real derivatives On harmonic functions Complex Series Series of complex terms Proposition 5..4: Linear combinations of series Proposition 5..8: Absolute convergence vs. convergence The comparison test d Alembert s ratio test The n th root test Power series Theorem 5.2.7: Hadamard s theorem The exponential, sine and cosine functions The exponential function Properties of the exponential function The complex conjugate and the modulus of the exponential function Euler s formula Trigonometric functions The sine and cosine functions of one complex variable Complex hyperbolic functions The real and imaginary parts Unboundedness Zeros RHS, version June 2, 28

3 Complex Analysis I CONTENTS 7 Complex mappings Some properties of the complex mappings Complex logarithms Argument Complex logarithms: the inverse of the exponential function Branches of the argument Branches of the complex logarithm General power function Integrals Complex-valued functions of real variables Curves and contours Integration along curves/contours Proposition 9.3.3: Invariance Proposition 9.3.5: The integral of the negative Proposition 9.3.6: Joining Integrals along curves with respect to the arclength Lemma 9.3.: Estimation lemma Theorems related to an antiderivative 69 Integral theorems 73 Theorem..: Goursat s theorem Theorem..5: The Cauchy Goursat theorem Cauchy s integral formulas 8 Theorem 2..: Cauchy s local integral formula Theorem 2..3: Cauchy s second formula Corollaries of the Cauchy integral formulas 85 Theorem 3..2: Morera s theorem Corollary 3..7: Liouville s theorem Theorem 3..: Identity theorem Theorem 3..7: Local maximum modulus theorem Theorem 3..2: Maximum modulus theorem Lemma 3..2: The Schwarz lemma Theorem 3..22: Minimum modulus theorem Global Cauchy theorem 9 4. Cycles Definition 4..2: Winding number Lemma 4..3: Winding number lemma Lemma 4..8: The component lemma Cauchy global integral theorem Theorem 4.2.: Cauchy global integral theorem Three corollaries of Cauchy global integral theorem Corollary 4.3.3: The case Ω = C {} Examples Deformation theorem Integrating rational functions on the real line The proof for the Cauchy global integral formula The extended complex plane The Riemann sphere and the extended complex plane Circlines The extended complex plane Behavior of functions at Möbius transformations 2 RHS, version June 2, 28 2

4 Complex Analysis I CONTENTS 6. The image of a circline is a circline Finding the image of a circline The image of a disc and the image of a half-plane Cross-ratios and the triplet representation of a Möbius transformation Möbius transformations preserve angles On conformal mappings Conformality at as analyticity at On conformal mapping problems References 9 Alphabetical Index RHS, version June 2, 28 3

5 Complex Analysis I BACKGROUND Background. The vector space R 2 We have studied the 2-dimensional vector space, R 2 = { x, x 2 ) x R, x 2 R }, where R is the set of real numbers. The vector space R 2 is equipped with two operations: addition between elements of R 2 and scalar multiplication between an element of R 2 and an element of R. That is, for every a = a, a 2 ) and b = b, b 2 ), and for λ R, a + b = a, a 2 ) + b, b 2 ) = a + b, a 2 + b 2 ) λa = λa, a 2 ) = λa, λa 2 ). Proposition... Recall that R 2, +) is an Abelian group. Proof. A) For all a, b R 2 we have a + b R 2. A2) For all a, b, c R 2, a + b) + c = a + b + c). A3) There exists a unique R 2, =, ), such that a + = + a = a for every a R 2. A4) For every a = a, a 2 ) R 2 there exists exactly one a R 2, a = a, a 2 ) = a, a 2 ) = a, a 2 ), such that a + a) = a + a =. A5) For all a, b R 2 : a + b = b + a. Remarks..2. A2) the operation + is associative. A3) the element is called zero element. A4) the element a is called the additive inverse of a A5) since the operator is commutative, R 2, +) is an Abelian group. b a y b a a + b = b + a x a b Figure.: Addition and subtraction in the vector space R 2, +) can be interpreted geometrically via the parallelogram law. Remark..3. The vector space R 2, +) is the Euclidean plane. Remark..4. Recall that R 2, +) is a real vector space, since R 2, +) is an Abelian group and the following statements hold: V) for all λ R, v R 2 λv R 2. V2) for all λ, μ R, v R 2 λμ)v = λμv). RHS, version June 2, 28 4

6 Complex Analysis I BACKGROUND V3) for the unit element R and every v R 2 v = v. V4) for all λ R 2 and for all v, u R 2 λv + u) = λv + λu. V5) for all λ, μ R and v R 2 λ + μ)v = λv + μv. Definition..5. Recall that we have introduced the inner product between a = a, a 2 ) R 2 and b = b, b 2 ) R 2 such that ab ) = a b + a 2 b 2. This is a mapping R 2 R 2 R. Another notation is ab. Remark..6. The notation a b is not used in this Complex Analysis course, since it is reserved for the multiplication which makes R 2, +, ) a field..2 The complex plane C We introduce multiplication between any two elements a and b from R 2 such that the multiplication takes as input a R 2 and b R 2 and gives as output c R 2. We define R 2 R 2 R 2 such that a, a 2 ) b, b 2 ) = a b a 2 b 2, a b 2 + a 2 b ) for every pair a = a, a 2 ) R 2 and b = b, b 2 ) R 2. In doing so we obtain that R 2, +, ) is a field. We write C = R 2, +, ) and call C the complex plane. Remark.2.. Here lies the difference between our vector calculus Vektorianalyysi II) course and this first complex analysis course. Remark.2.2. Multiplication is a little bit messy in Cartesian co-ordinates, but it becomes nicer in polar form, which we introduce later. This polar form gives a geometric meaning for multiplication of two vectors. In the vector calculus course we studied functions f R 2 R 2, where R 2 is the Euclidean plane. We looked at their differentiability and the properties of their integrals. Now we study functions f C C, where C = R 2, +, ) is the complex plane. It turns out that for a function f R 2 R 2 to have a derivative at every point in an open ball in R 2, that is, to be real differentiable in an open ball, is a weaker property than to have a complex derivative in every point of an open ball, that is, to be complex differentiable in an open ball. In summary: to be real differentiable in an open ball is easier than to be complex differentiable in the corresponding ball. Example.2.3. Let f R 2 R 2, x, y) x, y). The mapping f is real differentiable on the whole plane R 2, since its partial derivatives exist and are continuous. However, it turns out that is not complex differentiable. f C C, x, y) x, y), Before studying the complex differentiability and analyticity of functions, it will be useful to go through the elementary properties of the elements of C. Remark.2.4. We give an outline of why R 2, +, ) = C is a field. F) R 2, +) is an Abelian group. F2) R 2 {}, ) is an Abelian group: B) For all a R 2 and b R 2 : a b R 2. B2) For all a, b, c R 2 : a b) c = a b c). RHS, version June 2, 28 5

7 Complex Analysis I BACKGROUND B3) There exists, ) R 2 such that for every a R 2 : a, ) = a =, ) a. B4) For each a R 2 {}, a = a, a 2 ), there exists a unique inverse element of a, ) a a a 2 +, 2 a2 a = a R 2 a2 2 such that a a = a a =, ) =. B5) For all a, b R 2 : a b = b a. F3) For all a, b, c R 2 : a + b) c = a c + b c. This is called the st distributional law. Remarks.2.5. B2) The operation is associative. B3) The element, ) is called the unit element. Note: a, a 2 ), ) = a a 2, a + a 2 ) = a, a 2 ) and B5) yields, ) a, a 2 ) = a, a 2 ). B4) The point a is the inverse of a and we also write a. Let a = a, a 2 ) R 2 {}. We have to find x = x, x 2 ) R 2 {} such that x a = a x =, ), that is, x a = x, x 2 ) a, a 2 ) = x a x 2 a 2, x a 2 + x 2 a ) =, ). This holds if and only if { a x a 2 x 2 = and a x 2 + a 2 x = ; here a 2 + a2 2 >. There exists exactly one such x, x 2 ); since the determinant a a 2 = a a 2 a 2 + a2 2 ; the corresponding linear system has a unique solution: Remarks.2.6. x, x 2 ) = a a a 2 +, 2 a2 a a2 2. We sometimes write z z 2 = z z 2, where z C and z 2 C. 2. Let z C. We define z =, ) and for each n N: z n = z z n. Remark.2.7. The set { x, ) x R } C ). is a subfield of the field C. There is a mapping f R { x, ) x R }, which is a field isomorphism. This means that we can identify x R with x, ) C. Hence R C, that is, the set of real numbers is a proper subset of the set of complex numbers. Hence especially we can write, ) =. RHS, version June 2, 28 6

8 Complex Analysis I BACKGROUND Definition.2.8. The elements of C are called complex numbers. Definition.2.9. The complex number, ) C is called the imaginary unit and we write, ) = i. Remark.2.. Note that i 2 = i i =, ), ) =, + ) =, ) =, but we cannot take any roots, yet. We have not defined what the square root of z C is. RHS, version June 2, 28 7

9 Complex Analysis I 2 THE ELEMENTARY PROPERTIES OF COMPLEX NUMBERS 2 The elementary properties of complex numbers 2. Complex numbers can be written in Cartesian form Proposition 2... Every complex number z C can be expressed in a unique way z = x + iy, where x, y R. Proof. Let z C be fixed. Then z = x, y) for some x, y R. In particular, It is unique: z = x, y) = x, ) +, y) =, )x, ) +, )y, ) = x, ) + iy, ) = x + iy. Let z = x + iy = a + ib, where x, y, a, b R. Hence, By calculating the products we get x, ) +, )y, ) = a, ) +, )b, ). x, ) + y, + y) = a, ) + b, + b), which implies that x, y) = a, b). Thus x = a and y = b. Definition Let z = x + iy, where x, y R are fixed. The real part of z is and the imaginary part of z is x = Rez) R y = Imz) R. The set {z C Imz) = } is the real axis and the set {z C Rez) = } is the imaginary axis. The complex conjugate of z is defined by and the magnitude of z is z = x iy, z = x 2 + y 2. i Imaginary axis z z Real axis Figure 2.: Complex number z, its magnitudez and its complex conjugate z. Remark Two complex numbers are equal if and only if they have the same real part and the same imaginary part. Example z = 3 + i, z = 3 i, 2 z = 3) + 2 = 2, Rez) = 3, Imz) = z RHS, version June 2, 28 8

10 Complex Analysis I 2 THE ELEMENTARY PROPERTIES OF COMPLEX NUMBERS Imaginary axis i 3 + i 3 + i Real axis 3 i Figure 2.2: Complex number 3 + i from example 2..4, its magnitude 3 + i = 2 and its complex conjugate 3 i. Basic complex operations: Proposition For all z, z 2, z C: i) z + z 2 = z + z 2, ii) z z 2 = z z 2, iii) z = z. Proof. Let z = x + iy, z 2 = x 2 + iy 2 and z = x + iy, where x, x 2, y, y 2, x, y R. i) z + z 2 = x iy + x 2 iy 2 = x + x 2 ) iy + y 2 ) = z + z 2, ii) z z 2 = x, y )x 2, y 2 ) = x x 2 y y 2, x y 2 + x 2 y ) ) = z z 2, iii) z = x, y) = x, y) ) = x, y) = z. Note: given z = x + iy and z 2 = x 2 + iy 2, we have z + z 2 = x + x 2 ) + iy + y 2 ). The proofs to the following propositions are similar direct calculations as in the previous one, and are left as an exercise to the reader. Proposition For all z C: z = z, z =z, z z =z 2. Remark z z =z 2, thus if z C {}, then Proposition For every z C, z = z = z z z = z z. 2 Rez) = z + z) 2 and Imz) = z z). 2i RHS, version June 2, 28 9

11 Complex Analysis I 2 THE ELEMENTARY PROPERTIES OF COMPLEX NUMBERS The magnitudez behaves well with respect to the multiplication and division operations: Proposition z z 2 = z z 2, for all z, z 2 C, z = z z 2 z 2, for all z C and z 2 C {}, z n = zn, for all z C. With respect to addition and subtraction, we have the triangle inequality: Theorem 2..: The triangle inequality. More generally, Proof. z + z 2 z + z 2, for all z, z 2 C. z z 2 z + z 2 z + z 2, for all z, z 2 C. z + z 2 2 = z + z 2 ) z + z 2 ) = z + z 2 ) z + z 2 ) = z z + z z 2 + z z 2 + z 2 z 2 = z 2 + z z z 2 + z z 2 = z 2 + z Re z z 2 ) z 2 + z z z 2 = z + z 2 )2, which implies z + z 2 z + z 2. For the lower inequality, we have which implies Similarily, Note that z 2 z = z z 2, which gives us z = z z 2 + z 2 z z 2 + z 2, z z 2 z z 2, for all z, z 2 C. z 2 = z 2 z + z z 2 z + z. z 2 z z z 2, for all z, z 2 C. Hence we have shown that z z 2 z + z Complex numbers can be written in polar form Definition 2.2.: Polar coordinates. Let us recall the connection between Cartesian coordinates and polar coordinates from the vector calculus course. If the point x, y) R 2 {} is given, we can find r = x 2 + y 2 > and α [, 2π], counting counter-clockwise, such that r = x 2 + y 2, x = r cos α, y = r sin α. Conversely, if r > and α [, 2π] are given, we can find x and y. RHS, version June 2, 28

12 Complex Analysis I 2 THE ELEMENTARY PROPERTIES OF COMPLEX NUMBERS y x 2 + y 2 = r x, y) α x Figure 2.3: Polar coordinates r and α of vector x, y) R 2 {}. Let us remove the restriction on α. Let α R and note that { cos α = x r, sin α = y r. These equations determine α only up to an integer multiple of 2π. A complex number z C {} can be written in polar form as z = r cos α + ir sin α, 2.2.2) where r =z > is called the magnitude, modulus or absolute value of z and α R is called the argument or phase of z. The argument α is, however, only determined up to an integer multiple of 2π. Example z = 3 + i ) arg 3 + i = π + n2π, n Z 6 Im i z = 3 + i α = π 6 Re a) The complex number z and its argument. π 3 π 3 b) An equilateral triangle. Figure 2.4: The complex number z from example and an equilateral triangle, which can be used to find the argument of z. Note: The set of all possible arguments of z is denoted by Argz). Theorem 2.2.4: Euler s formula. The formula cos α + i sin α = e iα, α R, is called Euler s formula. We take it now as a notation, but we will prove it later The relationship between the Cartesian and the polar form is given by Euler s formula: z = x + iy = r cos α + ri sin α = rcos α + i sin α) = re iα. RHS, version June 2, 28

13 Complex Analysis I 2 THE ELEMENTARY PROPERTIES OF COMPLEX NUMBERS Remark If r e iα = r 2 e iα 2, then r = r 2 and α = α 2 + n2π for some n Z There are infinitely many choices for the argument of any given complex number. To get a unique argument we restrict the argument to lie in an interval α, β], where β α = 2π. The standard argument Arg π,π] is defined as the unique argument, of a complex number z, that lies in the interval π, π]. For the general interval α, β], where β α = 2π, the argument Arg α,β] is defined as the unique argument of z that lies in the interval α, β] Recall the formulas for the functions sine and cosine when α, β R: sin α + β) = sin α cos β + cos α sin β, cos α + β) = cos α cos β sin α sin β The geometric meaning of the multiplication of complex numbers For two complex numbers z = r e iα and z 2 = r 2 e iα 2, by Euler s formula, the definition of multiplication and we get z z 2 = r r 2 e iα e iα 2 = r r 2 cos α + i sin α )cos α 2 + i sin α 2 ) = r r 2 cos α cos α 2 sin α sin α 2 ) + icos α sin α 2 + sin α cos α 2 ) ) = r r 2 cos α + α 2 ) + i sin α + α 2 ) ) = r r 2 e iα +α 2 ). Notice the geometric meaning: when we multiply two complex numbers, the output is a complex number whose magnitude we obtain by multiplying the magnitudes of the input complex numbers. Likewise, the argument of the output complex number is obtained by summing the arguments of the input complex numbers. If z C {} is a fixed complex number, then in the mapping z z z, the point z is the input and the output is z z = z z e iarg z +Arg z). This means stretching and rotating in the complex plane C. Example Let f C C, z iz. The mapping f corresponds to a 9 degree rotation counterclockwise We have proved where As well as z z 2 = r r 2 e iφ +φ 2 ), z z 2 = z z 2, for all z, z 2 C, z = r e iφ, z 2 = r 2 e iφ 2, r j >, φ j R, j =, 2. Arg z z 2 ) = Arg z ) + arg z2 ), for all z, z 2 C {}. Remark Let α R and z, z 2 C {}. It might be the case that Example. Let z = i = z 2. Then Arg α,α+2π] z z 2 ) Arg α,α+2π] z ) + Arg α,α+2π] z 2 ). Arg π,π] z ) = Arg π,π] z 2 ) = π 2 and Hence, Arg π,π] z z 2 ) = Arg π,π] ) = π. Arg π,π] z z 2 ) = π π 2 + π ) = π = Arg 2 π,π] z ) + Arg π,π] z 2 ). RHS, version June 2, 28 2

14 Complex Analysis I 2 THE ELEMENTARY PROPERTIES OF COMPLEX NUMBERS i Im π π 2 Re i Figure 2.5: About Remark 2.2.2: z = i = z 2 and Arg π,π] z ) = Arg π,π] z 2 ) = π 2. z z 2 = i) i) = but Arg π,π] z z 2 ) = π Arg π,π] z )+Arg π,π] z 2 ) = π 2 + π. Proposition For all z C and z 2 C {} z = z z 2 z. 2 Furthermore, if z C {}, then ) z arg = argz ) argz 2 ). z 2 π 2 ) = Proof. Since we obtain Also, z = z z z z 2 = 2 z z 2, 2 z = z z 2 z 2. ) z argz ) = arg z z 2 = arg 2 z z 2 ) + argz 2 ). Remark Let α R and z, z 2 C {}. It might be the case that ) z Arg α,α+2π] Arg z α,α+2π] z ) Arg α,α+2π] z 2 ). 2 Example. Let z = and z 2 = 2. Then, Arg π,π] z z 2 Arg π,π] z ) =, Arg π,π] z 2 ) = π ) = π. ) = Arg π,π] 2 and Hence, Arg π,π] ) = π π = Arg 2 π,π] z ) Arg π,π] z 2 ). RHS, version June 2, 28 3

15 Complex Analysis I 2 THE ELEMENTARY PROPERTIES OF COMPLEX NUMBERS Im i 2 π Re Figure 2.6: About Remark 2.2.4: z = and z 2 = 2, so Arg π,π] z ) = and Arg π,π] z 2 ) = ) π. So z = z 2 2, but Arg π,π] = π π = Arg 2 π,π] z ) Arg π,π] z 2 ). Theorem 2.2.5: De Moivre s theorem. Let α R. Then for any n Z cos α + i sin α ) n = cos nα + i sin nα. Proof. Homework Applications of De Moivre s theorem. Homework.5.2) 2. Trigonometric formulas: Example. Proof. cos 2β = 2 cos 2 β, β R. cos 2β = Re cos 2β + i sin 2β ) cos ) ) 2 = Re β + i sin β ) = Re cos 2 β sin 2 β + i2 cos β sin β = cos 2 β sin 2 β ) = cos 2 β cos 2 β = 2 cos 2 β. 3. The study of the n th roots of complex numbers 2.3 The n th roots of complex numbers Let n 2, n N. A complex number z C {} is called an n th root of a complex number a C {}, if z n = a. Proposition Let If a =a cos θ + i sin θ ). n N, n 2, RHS, version June 2, 28 4

16 Complex Analysis I 2 THE ELEMENTARY PROPERTIES OF COMPLEX NUMBERS then a has the n distinct n th roots Remarks. z k =a n θ + k2π cos + i sin n ) θ + k2π, k =,,, n. n. Each of the n th roots of a has modulusa n. Hence, all the n th roots of a lie on a circle of radiusa n in the complex plane. 2. Since the argument of each successive n th root exceeds the argument of the previous root by 2π n, the nth roots are equally spaced on this circle. Proof. Let Our goal is to find all complex numbers a =a cos θ + i sin θ ). z =z cos φ + i sin φ ) such that De Moivre s theorem implies that z n = a. a cos θ + i sin θ ) =z n cos φ + i sin φ ) n =z n cos nφ + i sin nφ ). Hence, and we obtain that is z n cos nφ sin nφ = a = cos θ = sin θ z =a n nφ = θ + k2π, k Z. z =a n φ = θ+k2π, k Z. n Since by choosing k =,,, n, we obtain different values for z, we have ) z k =a θ + k2π θ + k2π n cos + i sin, where k =,,, n. n n On the other hand, the numbers z k satisfy the equation z n k = a. Example Find the fourth roots of a =. That is, solve the equation Solution. Since a = = cos π + i sin π, z k = cos π + k2π 4 + i sin z 4 =, z 4 + =. π + k2π, where k =,, 2, 3. 4 RHS, version June 2, 28 5

17 Complex Analysis I 2 THE ELEMENTARY PROPERTIES OF COMPLEX NUMBERS That is, z = cos π 4 + i sin π 4 = + i 2, z = cos 3π 4 + i sin 3π 4 z 2 = cos 5π 4 + i sin 5π 4 = + i 2, = i 2 and z 3 = cos 7π 4 + i sin 7π 4 = i 2. i Im z z Re z 2 z 3 Figure 2.7: The four fourth roots of a = Special case: the n th roots of unity The solutions of the equation are called the n th roots of unity and are given by z k = cos k2π n z n =, n N {}, k2π + i sin n = k2π ei n, k =,,, n. The roots represent the n vertices of a regular polygon of n sides inscribed in a circle of radius with center at the origin. Example Find fourth roots of unity. Solution. We notice that z 4 = z 4 = z 4 =. ) ) z 2 z 2 + = z ) z + ) z i) z + i). So the equation becomes z ) z + ) z i) z + i) =. And the four fourth roots of unity are, i, and i. RHS, version June 2, 28 6

18 Complex Analysis I 2 THE ELEMENTARY PROPERTIES OF COMPLEX NUMBERS i z Im z 2 z Re z 3 Figure 2.8: The four fourth roots of unity. RHS, version June 2, 28 7

19 Complex Analysis I 2 THE ELEMENTARY PROPERTIES OF COMPLEX NUMBERS 2.4 About history We mention only some names from the history of complex numbers and complex analysis. As a further reference, we refer to MacTutor, a web page by the University of St Andrews: Girolamo Gardano 5-576) Mentioned the square roots of negative numbers in his book Ars Magna, 545, but it seems that he thought they were useless. Raphael Bombelli ) Introduced the algebraic operations with complex numbers and used them to solve the 3 rd degree equations, L Algebra, 572. Gottfried Wilhelm Leibnitz ) Seemed to wonder if complex numbers are useful. Leonhard Euler ) Started to use complex numbers actively. For example, Euler studied analytic functions of one complex variable. Caspar Wessel ) A Norwegian-Danish engineer. The geometrical interpretation of complex numbers as vectors in the plane appeared in his writing from 799. Jean Robert Argand ) A French accountant, bookkeeper and an amateur mathematician. The geometric presentation of complex numbers as vectors in the plane is called Argand s diagram, although his publication is from 86. Carl Friedrich Gauss ) Started to study complex numbers systematically and to use their geometric properties. William Rowan Hamilton ) Introduced the concept of R 2, +, ). The complex function theory was developed in about 6 years by the following mathematicians. The properties of complex functions based on differentiation were developed by Bernhard Riemann ), integration were developed by Augustin-Louis Cauchy ), the power series were developed by Karl Theodor Wilhelm Weierstrass ). To summarise, complex numbers were discovered in the 5 s. However, the complex analysis was developed in about 6 years during the 8 s. RHS, version June 2, 28 8

20 Complex Analysis I 3 TOPOLOGICAL NOTIONS OF SETS IN THE COMPLEX PLANE 3 Topological notions of sets in the complex plane We give some simple topological properties which are necessary in our study of functions. The norm C C [, ), defines a metric or a distance function) d C C [, ), d z, z 2 ) = z z 2. Namely, for all z, z 2, z 3 C the following statements are valid:. z z 2 z z 3 + z 3 z z z 2 = z 2 z. 3. z z 2 =, if and only if z = z 2. The non-negative real number z z 2 is the distance between z, z 2 C. The space C, ) is a normed space and C, d), where d z, z 2 ) = z z 2 is defined by the norm, is a complete metric space. 3. Sets in the complex plane Let a C and r >. The open disc, Da, r), of radius r that is centered at a is defined by Da, r) = {z C z a < r}. The closed disc, Da, r), of radius r that is centered at a is defined by Da, r) = {z C z a r}. The boundary of both Da, r) and Da, r) is the circle C r a) = {z C z a = r}. The unit disc is the disc D, ). a r a r a) The open disc Da, r). b) The closed disc Da, r). Figure 3.: Open and closed disc. Definition 3... Let S be a set in C. A point z is an interior point of S if there exists a real number r >, such that D z, r ) S. The interior of S, denoted by int S, consists of all of its interior points. That is, int S = {z C z is an interior point of S}. RHS, version June 2, 28 9

21 Complex Analysis I 3 TOPOLOGICAL NOTIONS OF SETS IN THE COMPLEX PLANE Definition A set S is open, if every point contained in S is an interior point of S. A set S is closed, if its complement, C S, is open. S z r Figure 3.2: Let S be a set in C. A point z is an interior point of S if there exists r > such that D z, r ) S. Example 3..3 Examples of open sets).. D z, r ). 2. The punctured open disc, D z, r ) {z }. 3. The punctured complex plane, C {}. 4. The upper half-plane, H + = {z C Imz) > }. Example 3..4 Examples of closed sets).. D z, r ). 2. The circle, C r a). Definitions A point z C is a boundary point of S if every disc D z, r ) contains points belonging to both S and its complement, C S. The boundary of S is defined by S = {z C z is a boundary point of S}. A point z C is an accumulation point of S, if each punctured disc D z, r ) {z } contains at least one point of S. A point which is not an accumulation point is called an isolated point of S. The closure of S, denoted by S, is the union of S and its accumulation points. The open set U C is a neighbourhood of z, if z U. For example, D z, r ) is a neighbourhood of z. Remark. Open sets are useful for differentiation! Definition A set S is called bounded if there exist a real constant M >, such thatz < M whenever z S. In other words, the set S is contained in some large disc. If S is bounded, we define its diameter by diams) = sup z z 2, z, z 2 S. Conversely, a set is called unbounded, if it s not bounded. A set S is compact, if S is both closed and bounded. RHS, version June 2, 28 2

22 Complex Analysis I 3 TOPOLOGICAL NOTIONS OF SETS IN THE COMPLEX PLANE S D, M ) Figure 3.3: A compact set S in C Connectedness Recall that a continuous map, is a path in S. We write γ [a, b] S, a < b, γt) = γ t), γ 2 t)) = γ t) + iγ 2 t) The path γ joins the points γa) and γb). = Re γt) ) + i Im γt) ), t [a, b]. Definition. Let z, z 2 be given. The image of the path γt) = z + tz 2 z ), where t [, ], is a straight line segment: γ[, ]) = [γ), γ)] = [z, z 2 ]. Definition. A finite number of line segments joined end-to-end forms a polygonal line or a polygonal route. So if we have points z,, z n C, we call the union a polygonal route from z to z n. [z, z ] [z, z 2 ] [z n, z n ] RHS, version June 2, 28 2

23 Complex Analysis I 3 TOPOLOGICAL NOTIONS OF SETS IN THE COMPLEX PLANE γ a b a) A continuous mapping γ [a, b] S, where a < b, is a path in S. z 2 z z z 4 z 2 z 3 z b) A line segment in C. c) A polygonal route in C. Figure 3.4: A continuous mapping, a line segment and a polygonal route. Definition. A non-empty subset S of C is polygonally connected if, given two points z, z 2 S, there is a polygonal route from z to z 2 that lies entirely in S. Definition. An open set S is connected if each pair of points z, z 2 S can be joined by a polygonal line or route that lies entirely in S. S z 2 z Figure 3.5: A polygonally connected open set S. Remark. Connected sets are good for integration! Definition Domains). Open sets which are connected are called domains. RHS, version June 2, 28 22

24 Complex Analysis I 3 TOPOLOGICAL NOTIONS OF SETS IN THE COMPLEX PLANE 3.2 Limits and continuity 3.2. Sequences A complex sequence z n ) is an assignment of a complex number z n to each number n N. We write z n ) n= or just z n ). Definitions. A sequence z n ) is bounded if there exists a constant M R, such that z n M for all n. The sequence z n ) converges to a limit a C, written as z n a or lim n z n = a, if, given ε >, there exists a number N ε N, such that z n a < ε whenever n N ε. The sequence ω k ) is a subsequence of the sequence z n ), if there exist natural numbers n < n 2 <, such that ω k = z nk, for k =, 2,. Example. Let a C be given anda. Let z n ) be a sequence where z n = a n. Now, ifa <, then z n =an, which converges to. Ifa >, thena n and hence the sequence z n ) has no limit in C Limits of functions Let f S C be a function on a set S C. Let a be an accumulation point of S. Then or lim fz) = ω z a lim z a fz) = ω, z S that is, fz) ω as z a, if, given ε >, there exists δ = δ a,ε > such that whenever z S and <z a < δ. fz) ω < ε, Remark. The limit, if it exists, is determined by the behaviour of fz) as z approaches a. The value of fa) is irrelevant and it may not even be defined if a S. Example. Let fz) = Imz), z. Rez) Now we have Hence, lim z fz) fails to exist. fz) =, when z R {} fz) =, when z is on the line y = x Continuity Let f S C be a function. Then f is continuous at a S if, given ε >, there exists δ = δ a,ε >, such that fz) fa) < ε whenever a S andz a < δ. The function f is continuous on S if it is continuous at each a S. RHS, version June 2, 28 23

25 Complex Analysis I 3 TOPOLOGICAL NOTIONS OF SETS IN THE COMPLEX PLANE S fz) fa) ε z a δ C Figure 3.6: A function f S C is continuous at a S if, given ε >, there exists δ >, such that fz) fa) < ε whenever a S andz a < δ. Remarks. Sums and products of continuous functions are also continuous. Since the notions of convergence for complex numbers and points in R 2 are the same, the function f of a complex variable z = x+iy is continuous if and only if it is also continuous when viewed as a function of two real variables, x and y. If f is continuous, then the real-valued function is continuous. f C R, f z) = fz) The following lemma provides a link between convergence in C and R. Lemma Let z n ) be a complex sequence. Then z n ) converges, if and only if the real-valued sequences Rez n )) and Imz n )) both converge. If z n a, then z n a and z n a. ) Example. Let z n = + + i m Then z n + i. m ), m N. Definition Let f S C, f = u + iv, where u, v C R. So u = Re f and v = Im f. Now we define the functions u = Re f C R and v = Im f C R by setting Then, fz) ω implies We also have fz) ω and fz) ω. uz) = Re f)z) = Refz)), vz) = Im f)z) = Imfz)). Re fz) Re ω Im fz) Im ω. Proposition Let f S C. Then f is continuous at a S or on S), if and only if both Re f and Im f are continuous at a S or on S). 3.3 The Cauchy convergence principle for C Definition. A sequence z n ) is called a Cauchy sequence, or simply Cauchy, if z n z m, as n, m. RHS, version June 2, 28 24

26 Complex Analysis I 3 TOPOLOGICAL NOTIONS OF SETS IN THE COMPLEX PLANE Recall that R is complete: every Cauchy sequence of real numbers converges to a real number. Since the complex sequence z n ) is Cauchy if and only if Rez n )) and Imz n )) are, then we can conclude that every Cauchy sequence in C has a limit in C. Theorem. C is complete. RHS, version June 2, 28 25

27 Complex Analysis I 4 ANALYTIC FUNCTIONS 4 Analytic functions 4. Definitions Let Ω be an open set in C and f a complex-valued function on Ω we assume that f is a complexvalued function of one complex variable and Ω ). Definition. The function f is complex differentiable at the point z C if the quotient fz + h) fz ) h 4..) converges to a limit when h, where h C {} and z + h Ω. Recall C = R 2, +, ) is a field, and hence the quotient is well defined. The limit of 4.., when it exists, is denoted by f z ) and is called the complex derivative of f at z : f fz z ) = lim + h) fz ). 4..2) h h Note that h C {} must approach from any direction. Remark. Complex differentiability which is defined only at one point or two points is not enough to build an interesting, meaningful theory, so we need complex differentiability to be defined also on a small disc D z, ε ), ε >. Definition The function f is said to be analytic at z if there exists a disc D z, r ) such that f is complex differentiable at every point in the disc. z r Figure 4.: Open disc D z, r ). Remark. In order for f to be analytic at z, f should have a complex derivative at every point in a small neighbourhood of z. Definition Let Ω be an open set in C and f a complex-valued function on Ω. The function f is analytic on Ω if it is analytic at every point of Ω. Remark. Thus, analyticity is a stronger condition than differentiability. Now, we give some examples. Examples Any constant function is analytic on the whole complex plane. Let fz) = c C for all z C. Then f z) = : fz + h) fz) c c = h h =. 2. The function fz) = z is analytic on any open set Ω in C and f z) = : fz + h) fz) h = z + h z h =. RHS, version June 2, 28 26

28 Complex Analysis I 4 ANALYTIC FUNCTIONS 3. The function fz) = z 2 is analytic on any open set Ω in C and f z) = 2z: fz + h) fz) h = z + h)2 z 2 h = z2 + 2zh z 2 h Example The function fz) = z is not analytic. We have: fz + h) fz) h Hence there is no limit as h a. = z + h z h = 2z + h h 2z. = h h = { if h R {} if Reh) =, h. Note that, in terms of real variables, the function fz) = z corresponds to the map g x, y) x, y), which is differentiable in the real sense. So, the existence of the real derivative need not guarantee that f is analytic. Remark. The adjectives regular and holomorphic are sometimes used instead of analytic. Definition If a function is analytic on the entire complex plane, it is said to be entire. Entire functions are the very best class of complex functions. T. Tao) Remark. Note that, if D is a domain, then f is analytic if and only if f is complex differentiable at every point in D. 4.2 Elementary properties Lemma A function f is complex differentiable at z Ω if and only if there exists a complex number a such that fz + h) fz ) ah = hφh), 4.2.2) where φ is a function defined for all h, when h is very small and lim h φh) =. Here, a = f z ). Remark. The function φ depends on f and z, that is, φ = φ f,z. Proof. : Let us define φ) =, and φh) = fz +h) fz ) h : From 4.2.2), fz +h) fz ) h = a + φh) a as h. f z ), h. Corollary Let Ω be an open set in C. Let f be a complex-valued function on Ω. If f is differentiable at z Ω, then f is continuous at the point z Ω. Proof. From 4.2.2), lim h fz + h) fz ) =, and this is enough since z Ω is an accumulation point of the open set Ω. Corollary An analytic function is continuous. Remark. There are continuous functions which are not analytic. Example. fz) = z is continuous on the whole plane but it is not analytic Differentiation formulas). Let Ω be an open set in C. Let f and g be complex-valued functions on Ω. If the complex derivatives of f and g exist at a point z Ω, then. the function f + g Ω C is complex differentiable at z and f + g) z ) = f z ) + g z ), RHS, version June 2, 28 27

29 Complex Analysis I 4 ANALYTIC FUNCTIONS 2. the function f g Ω C is complex differentiable at z and fg) z ) = f z )gz ) + fz )g z ). 3. when gz ), then the function f g is well defined in some disc D z, r ), r > and ) f z g ) = f z )gz ) fz )g z ). gz ) 2 Proof. The proof is left as homework to the reader. Use lemma 4.2. or the definition. Corollary Any polynomial pz) = a + a z + + a n z n, where n N = {, 2, }, a k C, k =,,, n and a n, is analytic on the entire plane, and p z) = a + + na n z n. Corollary A rational function pz), where pz) and qz) are polynomials, is analytic on qz) any open set in which qz) is never zero. Examples The function fz) = is analytic on any open set in C that does not contain the origin and z f z) = z The rational function fz) = is analytic on any open set in C that does not contain ±i. z 2 + Example When is the function gz) = z 2 analytic? Since fz) = z is not analytic in any open set and z = z2 when z, the function g is not analytic anywhere. z Remark. The function g is complex differentiable at the origin. g + h) g) h = h2 h = h h. Remark. Note that the corresponding map k x, y) x 2 + y 2 is differentiable in the real sense on the whole R 2. Theorem 4.2.: Chain rule. Let Ω and U be open sets in C. Let f be analytic on Ω and let g be analytic on U and fω) U. The composite function g f, given by g f)z) = gfz)), is analytic on Ω and for all z Ω, g f) z) = g fz)) f z). f Ω ) U f g Ω g f C Figure 4.2: Chain rule Theorem 4.2.): the composite function g f)z) = gfz)) is analytic on Ω if functions g and f are analytic on their respective domains. RHS, version June 2, 28 28

30 Complex Analysis I 4 ANALYTIC FUNCTIONS Theorem 4.2.: Derivative of inverse function. Let Ω be an open set in C and let f Ω C be differentiable at z Ω, with f z). If there is a neighbourhood U of the point w = fz) such that f has a continuous inverse function f on U, then f is differentiable at w and Proof. Since f has a derivative at z, we have where εh) when h. f ) w) = f z) = f f w). fz + h) fz) = f z)h + hεh), Let k be so small that w + k U. Then for every such k C there exists h C with f w + k) = z + h. Since f is continuous in U, so h when k. Hence limz + h) = lim f w + k) = f w) = h, k k when k. f w + k) f w) k z + h z = w + k w = h fz + h) fz) h = f z)h + hεh) = f z) + εh) f z), 4.3 The relationship between the complex and real derivatives The notion of complex differentiability differs from the notion of real differentiability of a function of two real variables. Let us associate to a complex valued function f = u + iv, of one complex variable, the mapping g R 2 R 2, gx, y) = ux, y), vx, y) ). Recall from the vector calculus course that the function g is differentiable at a point x, y ), if there exists an R-linear transformation L R 2 R 2 such that with εh) as h. Equivalently, we can write as h, where h = h, h 2 ) R 2 {}. g x, y ) + h ) gx, y ) = Lh + h εh), g x, y ) + h ) gx, y ) Lh, h Remark. For a complex number z = x+iy, we write fz) = ux, y)+ivx, y), where u R 2 R and v R 2 R. The R-linear transformation L is unique and is called the derivative of g at x, y ). If g is differentiable in the real sense, the partial derivatives of u and v exist, and the R-linear transformation is described in the standard basis {e, e 2 } = {, ),, )} of R 2 by the Jacobian matrix of g: g x, y) = u x v x u y v y L. RHS, version June 2, 28 29

31 Complex Analysis I 4 ANALYTIC FUNCTIONS The derivative in the real case is a matrix, while the complex derivative is a complex number f z ). Our goal is to find the relationship between these two notions Let us assume that f has a complex derivative at the point z. We consider the limit in 4..2), where h R, i.e. h = h + ih 2 with h 2 =. Let us write z = x + iy, z = x + iy, and fz) = fx, y). Then, x ) ) f ) 4..2 f, y + h f ) x, y z = lim h h f ) ) x + h, y f x, y = lim h = f x x, y ), where denotes the usual partial derivative in the first variable by the definition of the partial x derivatives. Next we consider the limit 4..2) when h = ih 2, so h =. We obtain x ) ) f ) f, y + h f ) x, y z = lim h h f ) ) x, y + h 2 f x, y = lim h 2 ih 2 = f ) x, y i y, where y is the partial derivative in the 2nd variable. Recall that i =, ) and so ih 2 =, h 2 ) Hence, if f is analytic, we have shown that f x = f i y. h Writing f = u + iv, we find, after separating the real and imaginary parts and writing i that the partial derivatives of u and v exist. Moreover, they satisfy the equations = i, f x = x i f y which gives us the following identities: u v u + iv) = + i x x u = i u + iv) = i y y + v y = v u i y y, u x = v y and v x = u y ) These are called the Cauchy Riemann C R) equations. The Cauchy Riemann equations link together real and complex analysis Let us define two differential operators: and z = ) i 2 x y z = 2 x ) = ) + i. i y 2 x y RHS, version June 2, 28 3

32 Complex Analysis I 4 ANALYTIC FUNCTIONS Theorem If f is analytic at z, then and f z z ) =, 4.3.6) f z ) = f z z ) = 2 u z z ) ) If we write gx, y) = fz), then g is differentiable in the sense of real variables and det g x, y ) = f z ) ) Proof. For 4.3.6): Taking the real and imaginary parts, we see that the Cauchy Riemann equations are equivalent to f z z ) =. The C R equations imply: f z = f 2 x = 2 ) f + i = u v + i y 2 x x u x v y + i v x + u y + i u y ) ) C R) =. ) v + i2 y Conversely, if then = f z = ) ) u 2 x v y + i v x + u, y u x = v y and v x = u y. For 4.3.7): f z ) = 2 f z ) + 2 f z ) 4.3.) = f 2 x z ) ) = f z z ). f i y z ) ) Furthermore, we have f z 4.3.4) = 2 = 2 C-R) = 2 = 2 2 f x ) f i y u v + i x x i u y u x u y u x i u y u i ) y + u x = 2 u z. ) ) v + i y ) RHS, version June 2, 28 3

33 Complex Analysis I 4 ANALYTIC FUNCTIONS We defined two differential operators Let s recall theorem 4.3.5: If f is analytic at z, then z = ) i 2 x y and z = ) + i. 2 x y Also, if we write f z z ) = and f z ) = f z z ) = 2 u z z ). gx, y) = fz), then g is differentiable in the sense of two real variables, and det g x, y ) = f z ) 2. =z Let us write z = x, y ). Now, our goal is to show that g is differentiable in the real sense. We have to show that there exists an R-linear transformation L g R R such that gx, y ) + H) gx, y ) = L g H + H εh), 4.3.9) where εh) as H, H = h, h 2 ) R2 and h + ih 2 C. Recall from the vector calculus course that the R-linear transformation L g is unique and is called the derivative of g at x, y ). Since f is complex differentiable at z and the C R equations are satisfied, we know that By the definition of complex differentiability, where lim h εh) =. f z ) = f x z ) = u v + i x x = v u i y y. fz + h) fz ) = f z ) h + hεh) ) = v u i h + hεh) = y v y y i u y ) h + ih 2 ) + hεh), Now we identify a complex number with the pair of its real and imaginary parts. Hence, by equation and calculations from the start of this section, gx, y ) + H) gx, y ) = gx, y ) + h, h 2 )) gx, y ) where εh) = ε h), εh 2 )) and H, ). = fz + h) fz ) = = u x u x v x i u y u y v y ) h + ih 2 ) + hεh) h h 2 + hεh) = L g h + hεh) = L g H + H εh), RHS, version June 2, 28 32

34 Complex Analysis I 4 ANALYTIC FUNCTIONS Note that where εh) as H. h εh) = H h ε h) h 2 ε 2 h), h 2 ε 2 h) + h 2 ε h)), H = εh) So g is differentiable in the sense of two real variables. We obtain det g x, y ) = u = v x y v u x y ) 2 ) 2 u x + u y = u x u i y The next theorem gives an important converse: 2 = 2 u 2 z = f z ) 2. Theorem Suppose that f = u + iv is a complex-valued function defined on an open set Ω C. If u and v are differentiable in the real sense at the point z Ω, and their partial derivatives satisfy the C R equations at point z, then f has a complex derivative at z and f z ) = u x z v ) + i x z ). Proof. Let z = x, y ) Ω and h = h + ih 2 = h, h 2 ). Since u and v are differentiable at z, we have and uz + h) uz ) = Duz )h + h ε h) = uz )h) + h ε h) = u x z )h + u y z )h 2 + h ε h) where ε k h) as h, k =, 2. vz + h) vz ) = Dvz )h + h ε 2 h) = vz )h) + h ε 2 h) = v x z )h + v y z )h 2 + h ε 2 h), Let us write εh) = ε h) + iε 2 h). We obtain using the complex functions fz + h) fz ) = uz + h) uz )) + ivz + h) vz )) ) ) = u x z )h + u y z )h 2 + i v x z )h + v y z )h 2 By the C R equations + h εh). fz + h) fz ) = = = u ) ) x z )h v x z )h 2 + i v x z )h + u x z )h 2 + h εh) ) v + i h x + ih 2 ) + h εh) ) h εh) h + h, u x u x + i v x h RHS, version June 2, 28 33

35 Complex Analysis I 4 ANALYTIC FUNCTIONS h εh) where ε =, if h, ε h ) = and ε h) as h. Hence f is complex differentiable at z and f z ) = u x z v ) + i x z ). Corollary The function f = u + iv is complex differentiable at z if and only if u and v are differentiable in the real sense at z and their partial derivatives satisfy the Cauchy Riemann equations at z. Proof. One implication is Theorem The other implication follows from the proof of Theorem Remark Recall the following sufficient condition for u R 2 R to be differentiable in the real sense: Let Ω R 2 be an open set and u Ω R. If the partial derivatives of u exist at every point in Ω and are continuous at every point in Ω, then u is differentiable on Ω. There is a weaker sufficient condition: If the partial derivatives of u exist in a small ball Bz, r) with some r > and are continuous at z, then u is differentiable at z. In real life the following corollary is very important: Corollary Let Ω be an open set in C and let f = u + iv be a complex-valued function on Ω. If u and v have continuous partial derivatives on Ω and satisfy the C R equations on Ω, then f is analytic on Ω. Examples Let fz) = 2xy + ix 2 + y 2 ). Now, let u R 2 R and v R 2 R be such that ux, y) = 2xy and vx, y) = x 2 + y 2. The partial derivatives of u and v are u x = u x = 2y, v x = v x = 2x, u y = u y = 2x v y = v y = 2y. and Now u x, u y, v x, v y exists and are continuous, thus we know that f is differentiable in the real sense of two variables. C R equations are { ux = v y and u y = v x. The only points where the C R equations are satisfied are the points where x =, i.e., the points on the imaginary axis. Hence, f is complex differentiable at these points and f iy) = u x iy) + iv x iy) = 2y. Therefore f fails to be analytic there does not exist an open disk where the function has complex derivatives). 2. Let fz) = z) 2. Is f analytic? Now, let u R 2 R, ux, y) = Re fx, y) = x 3 3xy 2 and v R 2 R, vx, y) = Im fx, y) = y 3 3x 2 y. Homework. RHS, version June 2, 28 34

36 Complex Analysis I 4 ANALYTIC FUNCTIONS Im Re Figure 4.3: Function f in example is complex differentiable at the imaginary axis, but it is not analytic anywhere as there does not exist an open disk where the function has complex derivatives. Remark Let fx + iy) = x y, where x, y R. Show that f satisfies the C R equations at the origin, but f is not complex differentiable at the origin. Remark Let fx + iy) = x 2 + y 2 ) + i2yx. Now the partial derivatives f x f y = 2x + i2y and = 2y + i2x satisfy the C R equations. Do as in example Remark We defined the operators z = ) i and 2 x y z = ) + i. 2 x y The notation for this is clear/reasonable if we write x = 2 z + z) and y = z z), 2i working formally using the chain rule, we have the following equations: f z = f x x z + f y y z = f 2 x + f 2i y = 2 f z = f x x z + f y y z = f 2 x f 2i y = 2 f x + i f x i f y f y ), ) On harmonic functions Recall the Laplace operator: Δ = 2 x y ) Let Ω be an open set and f Ω R be a C 2 -function a function which is twice continuously differentiable in the real sense, we write f C 2 Ω)). If Δfx, y) = for all x, y) Ω, then f is called harmonic. Using the C R equations we obtain the following property for analytic functions: Proposition The real part and the imaginary part of an analytic function are harmonic. Warning: We will use the fact that an analytic function is twice continuously differentiable in fact, it is f C Ω)), but we will prove this property later. RHS, version June 2, 28 35

37 Complex Analysis I 4 ANALYTIC FUNCTIONS Let Ω be an open set. Let f Ω C be analytic, and the real part of f, Re f Ω C, Re f)z) = Re fz). By the C R equations, we have: Δ Re f = 2 Re f x 2 = x = x x + 2 Re f y 2 Re f + y Im f y y y Re f Im f =. x Hence, Δ Re f = and Re f is harmonic. In the same way, Δ Im f =, and Im f is harmonic. There exists the following theorem, which we do not prove: Theorem. If u is harmonic on D z, r ) with r >, then there exists an analytic map f on D z, r ) such that u = Re f. Example. ux, y) = ax + by + c, where a, b R, is harmonic since Δux, y) =. Theorem Let f be analytic on a disc D. If Re f is constant in D, then f is constant in D. Proof. Let f = u + iv = Re f + i Im f. If u = Re f is constant, then u x = u y =. Since f is analytic, the C R equations imply that v x = v y =. From the vector calculus course we know that this means that f is constant. Theorem If f is analytic on a disc D and Im f or f or arg f is constant, then f is constant. Remark Different ways to show that f is not complex differentiable at z :. If f is discontinuous at z, then f is not complex differentiable at z. 2. If the limit for fz) fz ) fails to exist when z z z z, then f is not complex differentiable at z. For an example, see Homework if the C R equations are not satisfied at x, y ) = z, then f is not complex differentiable at z. Here f = u + iv. 4. If f is not differentiable in the sense of two real variables at x, y ), then f is not complex differentiable at z = x, y ). 5. Use the differentiation formulas for sum, multiplication and quotient. For an example, see Homework Let Ω R 2 be an open set. Consider f Ω R 2, a function of two variables. Recall that f is differentiable at the point z if there exists an R-linear mapping from R 2 to R 2 the derivative, or differential fz ) such that where ε as h and z, h R 2. fz + h) fz ) = fz )h + h εh), On the other hand, the fact that the corresponding function f Ω C, Ω C, is complex differentiable at z means that there exists a complex number α such that where lim h εh) =. fz + h) fz ) = α h + hεh), Hence, f is complex differentiable at z if and only if f is differentiable in the real sense, and the derivative fz ) is of the form fz )h = α h, with α C. But the R-linear mappings from R 2 to R 2 of this type are exactly the C-linear mappings. So we have: RHS, version June 2, 28 36