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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 CauchyRiemann 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 complexdifferentiable 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 CauchyRiemann 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 x u ( x, 0) u (0 , 0) x = lim x x x = 1 , v x (0 , 0) = lim x v ( x, 0) v (0 , 0) x = lim x x x = 1 , u y (0 , 0) = lim y u (0 , y ) u (0 , 0) y = lim y y y = 1 , v y (0 , 0) = lim y v (0 , y ) v (0 , 0) y = lim y y y = 1 . Clearly, the derivatives exist and satisfy the CauchyRiemann equations at z = 0 . (ii) To show that f ( z ) is not complexdifferentiable at z = 0, we want to calculate the derivative f (0) = lim z f ( z ) f (0) z = lim z f ( z ) z Letting z = r e i , i.e. approaching the origin from angle , we have that x = r cos and y = r sin . So the derivative is found by the limit f (0) = lim r f ( r e i ) r e i = lim r r 3 cos 3 (1+ i ) r 3 sin 3 (1 i ) r 2 r e i = lim r cos 3 (1 + i ) sin 3 (1 i ) e i The value of the limit depends on 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 CauchyRiemann 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 CauchyRiemann 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 i/ 4 .] SOLUTION : To verify CauchyRiemann, we will use a more convenient form of the conditions: f x = if y ....
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