My Math Forum Equate Real and Imaginary Parts of Expression

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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(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 04:30 AM.
March 9th, 2017, 01:29 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?
.

Last edited by skipjack; March 10th, 2017 at 04:32 AM.

 March 9th, 2017, 05:39 PM #3 Global Moderator   Joined: May 2007 Posts: 6,379 Thanks: 542 As written, the expressions are incorrect. For example sin(t-t)=sin0=0.
March 9th, 2017, 07:17 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 05:43 AM.

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