Analytic Functions. Differentiable Functions of a Complex Variable

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1 Analytic Functions Differentiable Functions of a Complex Variable In tis capter, we sall generalize te ideas for polynomials power series of a complex variable we developed in te previous capter to general functions of a complex variable Once we ave proved results to determine weter or not a function is analytic, we sall ten consider generalizations of some of te more common single variable functions wic are not polynomials - namely trigonometric functions exponential functions 1 Analyticity te Caucy-Riemann Equations Recall tat for a polynomial P(x, y) = u(x, y) + iv(x, y) wit complex coefficients, we sowed tat it was analytic, or differentiable as a function of a complex variable if only if P y = ip x, or equivalently if it satisfied te Caucy-Riemann equations u x = v y u y = v x In tis section, we sall sow tat tis result can be partially extended to any function of a complex variable 11 Determining weter a Function is Analytic First we sall sow tat if a function of a complex variable is differentiable, ten it must satisfy te Caucy-Riemann equations (so it is a necessary condition to satisfy CR) Proposition 11 If f = u + iv is differentiable at z, ten f x f y exist satisfy te CR equations ie or f y = if x u x = v y ; u y = v x Proof Before we prove te result, we make a couple of observations First note tat since we are assuming tat f is differentiable, exists regardless of te way approaces 0 Secondly, we observe tat by definition, f(x +, y) f(x, y) f x = f(x, y + ) f(x, y) f y = 1

2 2 Now if we take to be real, ten we ave = 0 f(x + iy + ) f(x, y) f((x + ) + iy) f(x, y) f(x +, y) f(x, y) = = = f x (z) Likewise, taking = iη purely imaginary, we ave 0 i f(x, y + η) f(x, y) = = f y η 0 iη i Since te it exists independent of direction, it follows tat f x = f y /i or f y = if x Te CR equations follow Unfortunately, just because te CR equations exist does not mean tat a function is differentiable as te following example sows Example 12 Sow tat te function f(x, y) = xy satisfies te CR equations but is not differentiable at (0, 0) Here we ave for real, f(0 + ) f(0, 0) f x (0, 0) = f(0 + ) f(0, 0) f x (0, 0) = = = for imaginary, so 0 = f y (0, 0) = if x = 0 However, if we approac along te line y = x, we ave f(0 + (1 + i)) f(0, 0) 2 0 = (1 + i) (1 + i) = ± i = 0 = 0 for real In particular, tis it doesn t exist so f is not differentiable As te last example illustrates, Satisfying te CR equations are not a sufficient condition for analyticity (unlike wit polynomials power series) Under certain stronger assumptions owever, te CR equations are enoug Proposition 13 Suppose f x f y exist in a neigbourood of z Ten if f x f y are continuous at z f y (z) = if x (z), ten f is differentiable at z

3 Proof We need to sow tat te it exists independent of ow approaces 0 We sall do tis by sowing tat = f x (z) = u x (z) + iv x (z) were f = u(x, y) + iv(x, y) Let = were ζ η are real numbers We consider u(x, y) v(x, y) separately Observe tat u(z + ) u(z) = u(x + ζ, y + η) u(x, y) = 1 ([u(x + ζ, y + η) u(x + ζ, y)] + [u(x + ζ, y) u(x, y)]) Note tat u(x+ζ, y+η) u(x+ζ, y) is te cange of te single variable function f(y) = u(x + ζ, y) over te interval [y, y + η] Terefore, since f y exists is continuous (by assumption), so is u y, so te mean value teorem for single real variable functions implies tere exists some number, say y + ϑ 1 η (for some 0 ϑ 1 1) in te interval [y, y + η] suc tat u(x + ζ, y + η) u(x + ζ, y) = ((y + η) y)u y (x + ζ, y + ϑ 1 η) We get similar results for [u(x+ζ, y) u(x, y)]), for te components of te function v(x, y) giving u(z + ) u(z) v(z + ) v(z) Terefore, we get u y(x + ζ, y + ϑ 1 η) + ζ u x(x + ϑ 2 ζ, y) v y(x + ζ, y + ϑ 3 η) + ζ v x(x + ϑ 4 ζ, y) ( ) u y (z 1 )+iv y (z 2 ) + ζ ( ) u x (z 3 )+iv x (z 4 ) were z k z for k = 1, 2, 3, 4 as 0 Next we observe tat since if x = f y, we ave f x if x + ζf x Using tese equalities, subtracting f x from f y + ζ f x 3

4 4 we get Now note tat for all ( ) u y (z 1 )+iv y (z 2 ) + ζ f x (z) ) ( u y (z 1 )+iv y (z 2 ) f y (z) ( ) ( η u x (z 3 )+iv x (z 4 ) f y+ ζ ) f x ) + ζ ( u x (z 3 )+iv x (z 4 ) f x (z) η, ζ 1 u y (z 1 ) + iv y (z 2 ) f y (z), u x (z 3 ) + iv x (z 4 ) f x (z) 0 as 0, so it follows tat or Tus f x (z) 0 f x (z) = f x (z), in particular, f(z) is differentiable at z Usually we consider a function to be differentiable if it is differentiable in some interval (we would not usually consider a function to be differentiable only at a point, see example below) Example 14 Consider f(x, y) = x 2 + y 2 Here we ave f x = 2x f y = 2y, so bot partial derivatives are continuous By te last result, tis means f x is differentiable everywere te CR equations are satisfied, so just at te point (0, 0) For tis reason, wen we define analyticity of a function at a point, rater tan just requiring differentiability at a point, we want it to describe te local beaviour at near te point Rg Definition 15 We say f is analytic at z if f is differentiable in a neigbourood of z Similarly, f is analytic on a set S if it is differentiable on some open set containing S Definition 16 We call a function wic is differentiable everywere an entire function

5 12 Generalizing Results from Real Variable Calculus Now we ave a formal definition for an analytic function, we sall consider a small number of results wic we can generalized from teir real counterparts We start by considering ow to differentiate an inverse function of a complex analytic function In order to do tis, we need te following definition Definition 17 Suppose tat S T are open sets tat f is 1 1 on S wit F(S) = T g is called te inverse function of f on T if f(g(z)) = z for z T g is said to be te inverse of f at a point z 0 if it is te inverse in some neigbourood of z 0 Proposition 18 Suppose tat g is te inverse of f at z 0 tat g is continuous at z 0 If f is differentiable at g(z 0 ) if f (g(z 0 )) 0, ten g is differentiable at z 0 g 1 (z 0 ) = f (g(z 0 )) Proof For any z z 0 in a neigbourood of z 0, we ave g(z) g(z 0 ) z z 0 = 1 f(g(z)) f(g(z 0 )) g(z) g(z 0 ) Since g is continuous at z 0, g(z) g(z 0 ) as z z 0, so by te differentiability of f g(z) g(z 0 ) 1 = z z 0 z z 0 f (g(z 0 )), since by definition f (g(z 0 )) = z z0 f(g(z)) f(g(z 0 )) z z 0 Oter results wic generalize from single variable are te following Proposition 19 If f = u+iv is analytic in a region u is constant, ten f is constant Proof Since u is constant, u x = u y = 0 Since f is analytic, it satisfies te CR equations, so it follows tat v x = v y = 0 Applying Proposition 317, it follows tat u v are constant ence f is constant Proposition 110 If f is analytic in a region D if f is constant on D, ten f is constant on D Proof If f = 0, te proof is obvious Else we ave Taking partial derivatives, we ave u 2 + v 2 = C 0 uu x + vv x = 0 5

6 6 uu y + vv y = 0 (using te cain rule) Using te Caucy Riemann equations, tis can be modified to get uu x vu y = 0 wic is equivalent to Taking te difference, we get vu x + uu y = 0 u 2 u x uvu y = 0 v 2 u x + vuu y = 0 (u 2 + v 2 )u x = 0 since u 2 + v 2 0, it follows tat u x = v y = 0 We get a similar result for u y v x, so te result follows 2 Generalizing Functions from Real Variables We now consider generalizing some of te functions we know from real variable calculus (by generalizing, we mean tat we define a function of a complex variable wic agrees wit te original real function wen evaluated at purely real numbers) 21 A Complex Exponential Function We want to define a generalization of te real exponential function to complex variables Specifically, we want to define a function f(z) satisfying (i) f(z 1 + z 2 ) = f(z 1 )f(z 2 ) for any z 1, z 2 C (ii) f(x) = e x for any real x (i) From te book, pages 41-42: Questions Suppose tat f(z) is a function wic satisfies tese two conditions Ten it follows tat f(z) = f(x + iy) = f(x)f(iy) = e x f(iy), so we need to determine wat conditions will be imposed on te purely imaginary part of a complex number Terefore, suppose tat f(iy) = A(y)+ ib(y), so we ave f(x + iy) = e x A(y)ie x B(y) In order for f to be analytic, it needs to satisfy te Caucy Riemann equations, so we must ave u x = v y u y = v x or e x A(y) = e x B (y) e x A (y) = e x B(y)

7 7 Tis means tat A (y) = A(y) A general solution to tis differential equation is A(y) = α cos (y) + β sin (y) for some real numbers α β so we get B(y) = A (y) = β cos (y) + α sin (y), f(z) = e x (α cos (y) + β sin (y)) + ie x ( β cos (y) + α sin (y)) = αe x (cos (y) + i sin (y)) + βe x (sin (y) i cos (y)) If f(z) agrees wit e x for real numbers, we must ave 1 = f(0) = α iβ, so α = 1 β = 0 Putting all tis togeter, we ave te following: Definition 21 We define te complex exponential function f(z) = e z as f(z) = f(x + iy) = e x (cos (y) + i sin (y)) Proposition 22 Te exponential function satisfies te following: (i) e z = e x (ii) e z 0 for any value of z (iii) e iy = cis(y) (iv) e z = α as infinitely many solutions for α 0 (v) (e z ) = e z Proof Most of tese results are fairly trivial to prove (i) e z = e x cos (y) + i sin (y)) = e x (cos 2 (y) + sin 2 (y)) = e x (ii) Tis simply follows because e x 0 for any x cos (y) + i sin (y)) 0 for any y (iii) e z = e x+iy = x x e iy = e x (cos (y) + i sin (y)) so e iy = cos (y) + i sin (y)) (iv) Suppose α is some non-zero complex number Ten in polar form, we ave α = re iϑ = rcis(ϑ) for some r > 0 some angle 0 ϑ < 2π It follows tat solutions to e z = e x e iy = α will be x = ln (r) y = ϑ + 2kπ for any integer k (v) Recall tat f (z) = f x (z), so (e z ) = (e x e iy ) = e x e iy = e z

8 8 22 Defining te Trigonometric Functions Now we ave defined te exponential function, we can define trigonometric functions First we observe tat for real values of y, we ave e iy = cos (y) + i sin (y)) e iy = cos (y) i sin (y)) Tus for real values of y, we can define sin (y) = 1 ) (e iy e iy 2i cos (y) = 1 2 (e iy + e iy ) To define complex trigonometric, we simply extend tese definitions to complex variables Specifically, we define tem as follows: Definition 23 We define te complex functions sin cos as sin (z) = 1 ) (e iz e iz 2i cos (z) = 1 2 (e iz + e iz ) Te complex trig functions sare many of te identities wit teir real counterparts We illustrate wit an example Example 24 We sall sow tat te indentity cos 2 (z) +sin 2 (z) = 1 olds for complex numbers Using te definitions, we ave [ ( )] 2 [ )] cos 2 (z) + sin 2 (z) = e iz + e iz + (e iz e iz 2 2i = 1 ) (e 2iz e 2iz 1 ) (e 2iz 2 + e 2iz = Of course, tey also ave many differences (most notably, complex trig functions are not bounded) Homework: (Pages 41 & 42) 2,4,6,8,11,13,14,16,17,19,21

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