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Unformatted text preview: Solutions Homework 8. (1) Prove that if z = x + iy and f ( z ) = p (  xy  ), then the real part and imaginary part of f satisfy the CauchyRiemann equations at z = 0, but f is not differentiable at z = 0. Solution: u ( x,y ) = p  xy  and v ( x,y ) = 0. Hence ∂v ∂x ( x,y ) = ∂v ∂y ( x,y ) = 0 for all ( x,y ). To find ∂u ∂x (0 , 0) we use the definition: ∂u ∂x (0 , 0) = lim h → ,h ∈ R u ( h, 0) u (0 , 0) h = 0 . Similarly ∂u ∂y (0 , 0) = 0. Hence the CauchyRiemann equations hold at z = 0. On the other hand f ( t (1 + i )) f (0) t (1 + i ) =  t  t (1 + i ) does not have a limit as t → 0 in R . Hence f (0) does not exist. (2) Let G ⊂ C be an open and connected set and let f : G → C be a holomorphic function such that f ( z ) = 0 for all z ∈ G . Prove that f is constant on G . (Hint: let S = { z ∈ G : f ( z ) = f ( z ) } for some fixed z ∈ G and show that S is open and closed. You can use from undergraduate analysis that if a real valued differentiableclosed....
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This note was uploaded on 02/05/2012 for the course MATH 703 taught by Professor Schep during the Fall '11 term at South Carolina.
 Fall '11
 Schep
 Equations

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