# Simple question

Discussion in 'Math' started by boks, Dec 13, 2008.

1. ### boks Thread Starter Active Member

Oct 10, 2008
218
0
Can somebody explain why

$\frac{-1}{(2 \pi )(2+in)}[e^{- \pi (2 + in)} - e^{\pi (2 + in)}]$

can be written

$\frac{(-1)^n}{2+in} \frac{e^{2 \pi} - e^{-2 \pi}}{2 \pi}$

where n is an integer?

2. ### studiot AAC Fanatic!

Nov 9, 2007
5,003
522
Rewrite the numerator as

{[exp(2π) . exp(inπ)] - [exp(-2π) . exp(-inπ)]}

Noting that (Eulers theorem)
exp (iπ) = cos (π) + isin (π) = 1
exp(-iπ) = cos (π) - isin (π) = 1

and substituting

The (-1)$^{n}$

comes from putting n into Eulers formula, the result is alternately +1 and -1

Sorry the π (pi) symbols are almost indistinguishable from the n.

3. ### boks Thread Starter Active Member

Oct 10, 2008
218
0
Great, thanks!

4. ### boks Thread Starter Active Member

Oct 10, 2008
218
0
$cos \pi$ = -1

5. ### studiot AAC Fanatic!

Nov 9, 2007
5,003
522
Damn spell checker again! I'll try not to scribble so quickly in future.
you're right, it = -1 but that does work out.

6. ### boks Thread Starter Active Member

Oct 10, 2008
218
0
I still don't understand why it's

$\frac{(-1)^n}{2+in} \frac{e^{2 \pi} - e^{-2 \pi}}{2 \pi}$

and not

$\frac{-(-1)^n}{2+in} \frac{e^{2 \pi} - e^{-2 \pi}}{2 \pi}$

7. ### studiot AAC Fanatic!

Nov 9, 2007
5,003
522
Don't forget I rearranged your orignal expression by taking the -1 insde the bracket. It is part of the numerator.

The last term in the third line should read exp minus (i n pi)

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