# hmwk1 - x y Since f is analytic we know that u and v have...

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Math 220A Complex Analysis Solutions to Homework #1 Prof: Lei Ni TA: Kevin McGown Conway, Page 4, Problem 2. Basic geometry in R 2 says that we have equality in the expression | z 1 + ··· + z n | ≤ | z 1 | + ··· + | z n | if and only if all the z i lie on the same ray with base point at the origin. This is true exactly when arg z k = arg z ` for all k, ` , and since arg z k - arg z ` = arg( z k /z ` ), this happens if and only if z k /z ` 0 for all k, ` . ± Conway, Page 6, Problem 5. For z = cis(2 π/n ) it is clear that we have z n - 1 = ( z - 1)( z n - 1 + ··· + z + 1) = 0 . Since n 2, we have 0 < 2 π/n < 2 π and therefore z 6 = 1. The result follows. ± Conway, Page 44, Problem 19. We write z = z + iy and f ( z ) = u ( x, y ) + iv
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Unformatted text preview: ( x, y ). Since f is analytic, we know that u and v have continuous partial derivatives and u x = v y , u y =-v x . If we deﬁne f * ( z ) = f ( z ), then we have f * ( z ) = u * ( x, y ) + iv * ( x, y ) with u * ( x, y ) = u ( x,-y ) and v * ( x, y ) =-v ( x,-y ). We compute u * x ( x, y ) = u x ( x,-y ) = v y ( x,-y ) , u * y ( x, y ) =-u y ( x,-y ) = v x ( x,-y ) , v * x ( x, y ) =-v x ( x,-y ) , v * y ( x, y ) = v y ( x,-y ) , and therefore u * x = v * y and u * y =-v * x . This shows that f * is analytic. ± 1...
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