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March 9th, 2017, 12:31 AM  #1 
Newbie Joined: Jan 2017 From: United States Posts: 4 Thanks: 0  Equate Real and Imaginary Parts of Expression
Let h = s + it, then (e^h)  1  h = (e^s *cos(t) 1 s)+ i(e^s*sin(t) t) =(e^s*cos(t1) + e^s s) + i(e^s*sin(tt) +t(e^s 1)) This means that (e^s cos(t) 1 s) =(e^s cos(t1) + e^s 1 s) , and (e^s sin(t) t)= (e^s sin(tt) +t(e^s1)) But why? How can we see this? Last edited by skipjack; March 10th, 2017 at 04:30 AM. 
March 9th, 2017, 01:29 AM  #2  
Senior Member Joined: Feb 2016 From: Australia Posts: 841 Thanks: 306 Math Focus: Yet to find out.  Quote:
Last edited by skipjack; March 10th, 2017 at 04:32 AM.  
March 9th, 2017, 05:39 PM  #3 
Global Moderator Joined: May 2007 Posts: 6,203 Thanks: 486 
As written, the expressions are incorrect. For example sin(tt)=sin0=0.

March 9th, 2017, 07:17 PM  #4  
Math Team Joined: May 2013 From: The Astral plane Posts: 1,515 Thanks: 576 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
e^h  1  h = (e^(s *cos(t))  1  s) + i(e^(s*sin(t))  t) or $\displaystyle e^h  1  h = (e^{s ~ \cos(t)} 1 s)+ i(e^{s ~ \sin(t)} t)$ But in order to compare reals to reals etc. you have to have something to compare it to. What is your other expression? Dan Addendum: Are you trying to say that $\displaystyle e^{s ~ \cos(t)} 1 s =  i(e^{s ~ \sin(t)} t)$ and find a solution for s and t? It can't be done without using a numerical technique. Last edited by skipjack; March 10th, 2017 at 05:43 AM.  

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equate, expression, imaginary, parts, real 
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