Why does this complex fraction simplify to that form for integer n?

[leopard]

Can somebody explain why

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

can be written

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

where n is an integer?

 

[Dick]

E.g. e^(pi*(2+i*n))=e^(2*pi)*e^(i*pi*n). e^(i*pi*n)=cos(pi*n)+i*sin(pi*n). sin(pi*n)=0. cos(pi*n)=(-1)^n. That sort of thing.

 

[leopard]

I still don't understand why it's

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

and not

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

 

[Dick]

I still don't understand why it's

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

and not

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

Because it's not. If you do it carefully you'll see that they used the (-1) to flip the order of e^(-2pi)-e^(2pi) into e^(2pi)-e^(-2pi).