Complex Analysis 0. Preliminaries and Notation For a complex number z = x + iy, we set x = Re z and y = Im z, its real and imaginary components. Any nonzero complex number z = x + iy can be written uniquely in the iθ 2 2 1 polar form z = re where r = z = x + y and θ = tan− (y/x). r is called the norm, modulus, or absolute value of z| and| θ is called the argument of z, written arg z. Notice p that arg z is well defined only up to multiples of 2π. iz iz Euler’s formulas e = cos(z)+ i sin(z) and e− = cos(z) i sin(z) give − iz iz iz iz e + e− e e− cos(z)= and sin(z)= − . 2 2i Complex numbers can sometimes be used an intermediate step to solve problems which are given entirely in terms of real numbers. In fact, if this were not true there would be little reason to define complex numbers. n 1 kπ n Example. Let n 2 be an integer. Show that − sin( )= − . ≥ k=1 n 2n 1 2πi n/4 πi/2 n Solution. Set ω := e n . Thus ω = e = i andQ ω = 1. kπi/n kπi/n k/2 k/2 k kπ e e ω ω− ω 1 sin = − = − = − n 2i 2i 2iωk/2 Therefore n 1 n 1 n 1 n 1 − − k k k kπ ω 1 k=1− (ω 1) k=1− (ω 1) sin = −k/2 = n 1 n 1 (1+2+−...+n 1)/2 = n 1 n 1 n(−n 1)/4 n 2iω 2 − i − ω − 2 − i − ω − kY=1 kY=1 Q Q n 1 k n 1 k n 1 k − (ω 1) − (ω 1) − (ω 1) = k=1 − = k=1 − = k=1 − 2n 1in 1(ωn/4)n 1 2n 1in 1in 1 2n 1( 1)n 1 −Q − − Q − − − Q − − − Since (ωk)n = 1 for each k, each factor in the numerator satisfies the equation (z +1)n = 1. n n 1 n n 2 In other words, they satisfy z + nz − + 2 z − + ... + nz = 0. Since none are zero, they satisfy p(z) := zn 1 + nzn 2 + n zn 3 + ... + n = 0. Therefore they are all the − − 2 − roots of the degree n 1 polynomial p(z). In the general, the product of all the roots of a degree d polynomial− is ( 1)d times the constant term. Thus in this case, the numerator n 1 − is ( 1) − n. − 1 1. Complex Differentiation f : C C. We→ shall assume that f is defined on a domain D which is open and path connected (meaning that any two points in D can be joined by a path within D.) Using real and imaginary components we can write z = x+iy and f(z)= u(x,y)+iv(x,y) for real-valued functions u and v, and in this way, when convenient, we can regard f as a function f(x,y)= u(x,y),v(x,y) : R2 R2. For example, when c = a+bi C the function z cz : C C corresponds to the function→ (x,y) (ax by,ay + bx)∈ : R2 R2 or equivalently,7→ → the function given by 7→ − → x a b x − . y 7→ b a y The (complex) derivative of f is defined by f(z) f(z0) f ′(z0):= lim − , z z0 z z → − 0 if the limit exists. The assumption that the limit exists implies that the same value is obtained if the limit is taken as z approaches z0 along any line. Approaching along the x-axis gives f(x + iy0) f(x0 + iy0) f ′(x0 + iy0)= lim − =(∂xf)(x0,y0)=(∂xu + i∂xv)(x0,y0). (1) x x0 (x + iy ) (x + iy ) → 0 − 0 0 Similarly approaching along the y-axis gives 1 f ′(x + iy )= (∂ u + i∂ v)(x ,y ). (2) 0 0 i y y 0 0 Equating real and imaginary parts in (1) and (2) gives Cauchy-Riemann Equations: ∂ u = ∂ v, ∂ v = ∂ u x y x − y Conversely, suppose f : C C is differentiable when regarded as a function R2 R2 and also satisfies the Cauchy-Riemann→ Equations. The Jacobian matrix for the derivative→ is given by u u Df = x y , v v x y which, upon substitution from the Cauchy-Riemann equations, becomes the matrix corre- sponding to z (u + iv )z. It follows that f is complex differentiable with derivative 7→ x x f ′(z)= ux + ivx. Summing up, we have: Theorem. Let f : C C be differentiable when regarded as a function from R2 R2. Then f is differentiable→ as a complex function if and only if it satisfies the Cauchy-Riemann→ equations. 2 A function which is differentiable at every point in a domain D is called holomorphic on D. In polar form, the CR-equations are as follows. Set x = r cos θ, y = r sin θ. Then ∂u ∂u ∂u ∂v ∂v 1 ∂v = cos θ + sin θ = cos θ sin θ = ∂r ∂x ∂y ∂y − ∂x r ∂θ and similarly ∂v 1 ∂u = . ∂r − r ∂θ Example. Let D = z C Re z > 0 . Recall that arg z is well defined only up to multiples of 2π. Define{ f ∈: D | C by f(z)} = log( z )+ i arg(z) where we choose the value of arg(z) which lies in ( π, π).→ | | − ∂u ∂ log(r) 1 ∂r 1 x 1 x x = = = = = ∂x ∂x r ∂x r x2 + y2 r r r2 p and 1 2 ∂v ∂θ ∂θ ∂ tan− (y/x) ( y/x ) y y = = = = − = − = − ∂x ∂x ∂x ∂x 1+(y/x)2 x2 + y2 r2 ∂u ∂v Similarly we can calculate ∂y and ∂y and verify that the Cauchy-Riemmann equations are satisfied. Therefore f(z) is differentiable. ∂f ∂u ∂v According to (1), ∂z = ∂x + i ∂x and so df x iy 1 1 = − = = dz r2 x + iy z A matrix of the form x a b x − . y 7→ b a y corresponds to the linear transformation consisting of the composition of rotation by 1 2 2 2 2 cos− a/(a + b ) and scalar multiplication by a + b . Thus the Jacobian matrix for Df corresponds to the composition of rotation by arg f ′(z) and multiplication by f ′(z) . | | 3 2. Complex Integration For a differentiable parameterized curve γ(t)= x(t)+iy(t):[a,b] C let γ′(t) C denote → ∈ the derivative γ′(t) := x′(t)+ iy′(t). As discussed in MATB42, γ′(t) gives the tangent vector to the curve γ(t), corresponding to the velocity at time t of a point moving along curve with position γ(t) at time t. Let f : B C be continuous where B is a domain → containing the curve γ. (That is, γ([a,b] B.) Assume that γ′(t) is a continuous function ⊂ of t. Define γ f(z) dz by R b f(z) dz := f γ(t) γ′(t) dt Zγ Za where the right hand side is defined by taking integrals of the real and imaginary compo- nents. In other words, b b f(z) dz := Re f γ(t) γ′(t) dt + i Im f γ(t) γ′(t) dt. γ a a Z Z Z More generally, for a piecewise differentiable curve γ, the integral can be defined by adding the integrals on the subintervals on which γ is differentiable. It is also possible to relax the condition that f be continuous, although we shall not need to consider such cases. As in line integrals in MATB42, the sign of the answer depends upon the orientation of the curve γ which is determined by the given parameterization. Example. Compute γ zdz where γ is the straight line joining 0 to 1+ i/2. Solution. ParameterizeR γ by γ(t)= t + it/2, 0 t 1. ≤ ≤ 1 1 1 1 zdz = (t+it/2)d(t+it/2) = (t+it/2)(1+i/2) dt = (1+i/2)2 tdt = (1+i/2)2. 2 Zγ Z0 Z0 Z0 2 Example. Compute C z dz where C is the unit circle, oriented counterclockwise. Solution. Parameterize C by C(t) = cos(t)+ i sin(t), 0 t 2π. R ≤ ≤ 2π 3 2 cos(t)+ i sin(t) t=2π z2 dz = cos(t)+ i sin(t) d cos(t)+ i sin(t) dt = = 0. C 0 3 t=0 Z Z 1 Example. Compute C z dz where C is the circle of radius R, oriented counterclockwise. Solution. Parameterize C by C(t)= R cos(t)+ iR sin(t), 0 t 2π. R ≤ ≤ 1 2π 1 2π dz = R sin(t)+ iR cos(t) dt = idt = 2πi. z R cos(t)+ iR sin(t) − ZC Z0 Z0 The calculation could equivalently be expressed by writing C(t)= Reit, 0 t 2π, giving ≤ ≤ 1 2π d(Reit) 2π dz = = idt = 2πi. z Reit ZC Z0 Z0 Note in particular that the answer is independent of the radius of C. 4 Proposition. For f = u + iv, f(z) dz = u(x,y) dx v(x,y) dy + i v(x,y) dx + u(x,y) dy − Zγ Zγ Zγ Proof. b f(z) dz = u x(t),y(t) + iv x(t),y(t) x′(t)+ iy′(t) dt Zγ Za b = u x(t),y(t) x′(t) v x(t),y(t) y′(t) − Za b + i v( x(t),y(t) x′(t)+ u x(t),y(t) y′(t) Za = u(x,y) dx v(x,y) dy + i v(x,y) dx + u(x,y) dy − Zγ Zγ Recall (MATB42) that a differential form ω is called closed if dω = 0.