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

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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 = 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 = x 2 + y 2 and v = 0. We find that ∂u ∂x = x x 2 + y 2 , ∂u ∂y = y 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 x 2 + y 2 = y 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 | ) = 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 0 C , z 0 = 0. Then ( f ( z )) 2 = | z | 2 = z z would be analytic at z 0 and, since z 0 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

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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 . (c) If f ( z ) = z = x - iy then ∂f ∂x = 1, ∂f ∂y = - i. Therefore ∂f ∂z = 1 2 ∂f ∂x + 1 i ∂f ∂y = 1 2 (1 - 1) = 0 and ∂f z = 1 2 ∂f ∂x - 1 i ∂f ∂y = 1 2 (1 + 1) = 1 .
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