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?

 

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.

 

Great, thanks!

 

\(cos \pi\) = -1

 

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

 

I still don't understand why it's

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

and not

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

 

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)