hw4_soln - Complex Analysis Fall 2007 Homework 4: Solutions...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Complex Analysis Fall 2007 Homework 4: Solutions 1.5.2. (a) The function f ( z ) = 3 z 2 +7 z +5 is a polynomial so is analytic everywhere with derivative f ( z ) = 6 z + 7. (b) The function f ( z ) = (2 z +3) 4 is a composition of polynomials so is analytic everywhere with derivative f ( z ) = 8(2 z + 3) 3 (by the chain rule). (c) The function f ( z ) = (3 z- 1) / (3- z ) is rational and so is analytic where z 6 = 3. By the quotient rule, its derivative is f ( z ) = 8 (3- z ) 2 . 1.5.10. Let f ( z ) = | z | and write f = u + iv and z = x + iy . Then u = p x 2 + y 2 and v = 0. We find that ∂u ∂x = x p x 2 + y 2 , ∂u ∂y = y p x 2 + y 2 ∂v ∂x = ∂v ∂y = 0 . In order for f to be analytic the Cauchy-Riemann equations must hold. That is, we need x p x 2 + y 2 = y p x 2 + y 2 = 0 which is impossible, since it requires x and y to be both simultaneously zero and nonzero. Therefore f is differentiable nowhere. Here’s another proof. Since the partial derivatives of Re( | z | ) = p x 2 + y 2 do not exist at (0 , 0), the Cauchy-Riemann theorem implies that | z | cannot be differentiable at z = 0. Suppose that f ( z ) = | z | were analytic at some z ∈ C , z 6 = 0. Then ( f ( z )) 2 = | z | 2 = z z would be analytic at z and, since z is nonzero, so too would be ( f ( z )) 2 /z = z . But we know that g ( z ) = z is analytic nowhere, so this is impossible. Therefore f ( z ) = | z | cannot be analytic at any nonzero complex number either. Hence, | z | is differentiable nowhere, i.e. is not analytic. 1.5.14. (a) According to Cauchy-Riemann equations, if f is analytic then f = ∂f ∂x = 1 i ∂f ∂y 1 so that ∂f ∂z = 1 2 ∂f ∂x + 1 i ∂f ∂y = 1 2 ( f + f ) = f . (b) If f ( z ) = z = x + iy then f is analytic so that by part (a) we have ∂fz = f = 1 . Moreover ∂f ∂x = 1, ∂f ∂y = i so that ∂f ∂ z = 1 2 ∂f ∂x- 1 i f y = 1 2 (1- 1) = 0 ....
View Full Document

This note was uploaded on 02/29/2012 for the course MATH 4364 taught by Professor Ryandaileda during the Fall '11 term at Trinity University.

Page1 / 6

hw4_soln - Complex Analysis Fall 2007 Homework 4: Solutions...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online