Excercise Sheet Solution 2

Excercise Sheet Solution 2 - MATH 20142 Complex Analysis...

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MATH 20142 Complex Analysis Solution Sheet 2 1. (a) For each z 0 C we have f ( z 0 ) = lim z z 0 z 2 + z ( z 2 0 + z 0 ) z z 0 = lim z z 0 ( z z 0 )( z + z 0 + 1) z z 0 = lim z z 0 ( z + z 0 + 1) = 2 z 0 + 1 , whence f ( z ) = 2 z + 1. (b) For z 0 ̸ = 0, f ( z 0 ) = lim z z 0 1 /z 2 1 /z 2 0 z z 0 = lim z z 0 z 2 0 z 2 z 2 0 z 2 ( z z 0 ) = lim z z 0 z 0 + z z 2 0 z 2 = 2 z 3 0 , whence f ( z ) = 2 /z 3 . (c) For each z 0 C we have f ( z 0 ) = lim z z 0 z 3 z 2 ( z 3 0 z 2 0 ) z z 0 = lim z z 0 ( z z 0 )( z 2 + z 0 z + z 2 0 z z 0 ) z z 0 = 3 z 2 0 2 z 0 , whence f ( z ) = 3 z 2 2 z . Note that the complex derivatives are identical to their real analogues. 2. (a) For z ̸ = 0, z + 1 z = x + iy + 1 x + iy = x + iy + x iy x 2 + y 2 . Therefore, u ( x, y ) = x + x x 2 + y 2 , v ( x, y ) = y y x 2 + y 2 for ( x, y ) ̸ = (0 , 0). Let us verify the Cauchy-Riemann equations. We have ∂u ∂x = 1 + x 2 + y 2 2 x 2 ( x 2 + y 2 ) 2 = 1 + y 2 x 2 ( x 2 + y 2 ) 2 , ∂v ∂y = 1 x 2 + y 2 2 y 2 ( x 2 + y 2 ) 2 = 1 + y 2 x 2 ( x 2 + y 2 ) 2 , 1
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i.e., ∂u/∂x ∂v/∂y . Similarly, ∂u/∂y ≡ − ∂v/∂x . [We omit the computation.]
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