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 March 8th, 2017, 11:31 PM #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(t-1) + e^s -s) + i(e^s*sin(t-t) +t(e^s -1)) This means that (e^s cos(t) -1 -s) =(e^s cos(t-1) + e^s -1 -s) , and (e^s sin(t) -t)= (e^s sin(t-t) +t(e^s-1)) But why? How can we see this? Last edited by skipjack; March 10th, 2017 at 03:30 AM. March 9th, 2017, 12:29 AM   #2
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Quote:
 Originally Posted by Robart Let $h = s + it$, then, \displaystyle \begin{align*} (e^h) - 1 - h &= (e^s * \cos(t) - 1 - s)+ i(e^s*\sin(t) - t) \\ &=(e^s * \cos(t - 1) + e^s - s) + i(e^s*\sin(t-t) +t(e^s -1)). \end{align*} This means that, $(e^s \cos(t) - 1 - s) = (e^s \cos(t-1) + e^s -1 -s)$, and $(e^s \sin(t) - t) = (e^s \sin(t - t) + t(e^s-1)).$ But why? How can we see this?
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Last edited by skipjack; March 10th, 2017 at 03:32 AM. March 9th, 2017, 04:39 PM #3 Global Moderator   Joined: May 2007 Posts: 6,806 Thanks: 716 As written, the expressions are incorrect. For example sin(t-t)=sin0=0. March 9th, 2017, 06:17 PM   #4
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Quote:
 Originally Posted by Robart Let h = s + it, then (e^h) - 1 - h = (e^s *cos(t) -1 -s)+ i(e^s*sin(t) -t)
With the addition of a few parentheses, this is correct. I don't know see how you got the line just below it. It should read:
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 04:43 AM. Tags equate, expression, imaginary, parts, real Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Haz5069 Complex Analysis 1 December 3rd, 2016 02:08 PM xamdarb Algebra 4 March 9th, 2014 06:13 PM consigliere Calculus 3 January 30th, 2014 06:06 AM szak1592 Complex Analysis 2 March 28th, 2013 09:47 AM ElusiveNeutrino Complex Analysis 2 March 23rd, 2009 07:44 AM

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