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118Homework2 - Homework 2 Due Wednesday Sept 8 1 Suppose...

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Unformatted text preview: Homework 2 Due Wednesday, Sept. 8 1. Suppose that f : C → C and g : C → C and that lim z → z f ( z ) = 0. Prove that if there are positive real numbers M and r such that | g ( z ) | ≤ M whenever | z- z | < r , then lim z → z f ( z ) g ( z ) = 0. (Note that if f ( z ) = z and g ( z ) = 1 /z , then lim z → f ( z ) = 0 but lim z → f ( z ) g ( z ) = 1 6 = 0. Why doesn’t this contradict the result you have proven above?) 2. Define f : C → C by f ( z ) = z 3 . (a) Use the definition of the derivative to calculate f ( z ). (b) Writing f ( z ) = u ( x, y ) + iv ( x, y ) verify that the Cauchy-Riemann equations hold, and that f ( z ) = u x ( x, y ) + iv x ( x, y ) = v y ( x, y )- iu y ( x, y ) . 3. Let g : C → C by g ( z ) = z 2 . Use the Cauchy-Riemann equations to show that for z 6 = 0 the derivative g ( z ) does not exist. Also prove that g is differentiable at 0. (Hint: Use the definition of g (0).) 4. For f : C → C with f = u + iv define the conjugate of f to be the function...
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118Homework2 - Homework 2 Due Wednesday Sept 8 1 Suppose...

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