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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 CauchyRiemann 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 CauchyRiemann 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 CauchyRiemann 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 ....
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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.
 Fall '11
 RyanDaileda
 Polynomials, Derivative

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