s3_95-100aSOLUTIONS

s3_95-100aSOLUTIONS - ACM95a/100a Last updated Problem Set...

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Unformatted text preview: ACM95a/100a Last updated October 20, 2009 Problem Set III Solutions Prepared by Faisal Amlani 1. (20 pts.) Consider the complex function f ( z ) = u + iv = x 3 (1+ i ) − y 3 (1 − i ) x 2 + y 2 for z negationslash = 0 for z = 0 Show that the partial derivatives of u and v with respect to x and y exist at z = 0 and that u x = v y and u y = − v x there: the Cauchy-Riemann equations are satisfied at z = 0. On the other hand, show that lim z → f ( z ) z does not exist, that is, f is not complex-differentiable at z = 0. Does this example contradict our theory? Explain. SOLUTION : (i) Separating the real and imaginary parts gives u = braceleftBigg x 3 − y 3 x 2 + y 2 for z negationslash = 0 , for z = 0 . , v = braceleftBigg x 3 + y 3 x 2 + y 2 for z negationslash = 0 , for z = 0 . Recall the Cauchy-Riemann equations: u x = v y , u y = − v x . The partial derivatives of u and v at z = 0 (i.e. the point x = y = 0): u x (0 , 0) = lim Δ x → u (Δ x, 0) − u (0 , 0) Δ x = lim Δ x → Δ x − Δ x = 1 , v x (0 , 0) = lim Δ x → v (Δ x, 0) − v (0 , 0) Δ x = lim Δ x → Δ x − Δ x = 1 , u y (0 , 0) = lim Δ y → u (0 , Δ y ) − u (0 , 0) Δ y = lim Δ y → − Δ y − Δ y = − 1 , v y (0 , 0) = lim Δ y → v (0 , Δ y ) − v (0 , 0) Δ y = lim Δ y → Δ y − Δ y = 1 . Clearly, the derivatives exist and satisfy the Cauchy-Riemann equations at z = 0 . (ii) To show that f ( z ) is not complex-differentiable at z = 0, we want to calculate the derivative f ′ (0) = lim Δ z → f (Δ z ) − f (0) Δ z = lim Δ z → f (Δ z ) Δ z Letting Δ z = Δ r e iθ , i.e. approaching the origin from angle θ , we have that x = Δ r cos θ and y = Δ r sin θ . So the derivative is found by the limit f ′ (0) = lim Δ r → f (Δ r e iθ ) Δ r e iθ = lim Δ r → Δ r 3 cos 3 θ (1+ i ) − Δ r 3 sin 3 θ (1 − i ) Δ r 2 Δ r e iθ = lim Δ r → cos 3 θ (1 + i ) − sin 3 θ (1 − i ) e iθ The value of the limit depends on θ and is not a constant in every direction. So the limit does not exist and we can conclude that the function is not differentiable at z = 0. This does not contradict our theory if we recall that satisfying the Cauchy-Riemann equations is a necessary but not sufficient condition for differentiability. 2. (20 pts.) Show that the function f ( z ) = e − z- 4 for z negationslash = 0 for z = 0 satisfies the Cauchy-Riemann equations everywhere, including at z = 0, but f ( z ) is not analytic at the origin. Does this example contradict our theory? Explain. [Hint: Consider the ray z = re iπ/ 4 .] SOLUTION : To verify Cauchy-Riemann, we will use a more convenient form of the conditions: f x = − if y ....
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This note was uploaded on 01/14/2010 for the course ACM 1 taught by Professor Prof during the Fall '09 term at Caltech.

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s3_95-100aSOLUTIONS - ACM95a/100a Last updated Problem Set...

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