My Math Forum Proving e^z is an entire function

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 February 25th, 2014, 03:53 PM #1 Senior Member   Joined: Jan 2014 Posts: 196 Thanks: 3 Proving e^z is an entire function The following asked to show that $e^z$ is an entire function. I am using the Cauchy-Rieman equations to prove this, and not understanding something. $e^z= e^x cosy + ie^x siny$ $\frac{\partial u }{\partial x}= e^x cosy = \frac{\partial v}{\partial y}$ but $\frac{\partial u}{\partial y}= -e^x siny \neq -\frac{\partial v}{\partial y} = -e^x cosy$ Am I missing something, doing something wrong? Thanks
 February 25th, 2014, 04:41 PM #2 Senior Member   Joined: Jan 2014 Posts: 196 Thanks: 3 Re: Proving e^z is an entire function I see my error. the Cauchy-Rieman equations were not correct in my first post. $\frac{\partial u}{\partial y}= -\frac{\partial v}{\partial x}$ which both equal $-e^x sin y$
 February 25th, 2014, 07:54 PM #3 Math Team     Joined: Aug 2012 From: Sana'a , Yemen Posts: 1,177 Thanks: 44 Math Focus: Theory of analytic functions Re: Proving e^z is an entire function I think you are making the same error in the other post ?

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