My Math Forum Proving e^z is an entire function

 Complex Analysis Complex Analysis Math Forum

 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 ?

 Tags entire, function, proving

,

,

,

,

,

,

,

,

,

,

,

,

,

,

# show the e rest to a is entire function

Click on a term to search for related topics.
 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post beanzzzz Calculus 2 January 12th, 2014 04:00 PM William_33 Complex Analysis 2 April 16th, 2013 11:59 AM snjvsingh Complex Analysis 1 April 20th, 2011 09:33 AM Erdos32212 Complex Analysis 1 April 18th, 2010 11:44 AM fed2black Calculus 1 November 19th, 2009 10:47 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top