Question: In complex analysis,I have meet this follow problem:

> 
> let $s$ is complex numbers,and $i^2=-1$
> 
> 
> prove or disprove
> $$i^{s-1}+(-i)^{s-1}=2\sin{\dfrac{\pi s}{2}}$$
> 
> 
> 

I know when $s$ is an integer,this is clear true,But for any complex numbers, I can't it.Thank you

This problem is from when I deal this: <http://www.maths.tcd.ie/pub/HistMath/People/Riemann/Zeta/EZeta.pdf>

Answer: 
Write $\pm i^\alpha = e^{\pm \alpha i(\pi/2+2\pi n)}$ for $n \in \mathbb{Z}$. Then 

\begin{align\*}
i^{s-1} + (-i)^{s-1} &= e^{i(s-1)(\pi/2+2\pi n)} + e^{-i(s-1)(\pi/2+2\pi n)} \\
&= 2\cos((s-1)(\pi/2+2\pi n))\\
&= 2\cos(s\pi/2+2s\pi n-\pi/2-2\pi n) \\
&= 2\sin(s\pi/2+2s\pi n-2\pi n) \\
\end{align\*}

This is as far as you can simplify unless you assume that you are using the principal value ($n = 0$).
Because

$$i=e^{i \pi/2},-i=e^{-i \pi/2}$$

So
$$i^{s-1}+(-i)^{s-1}=e^{i (s-1)\pi/2}+e^{-i (s-1)\pi/2}=2\cos\left((s-1)\frac{\pi}{2}\right)=2\cos\left((1-s)\frac{\pi}{2}\right)=2\sin\left(\frac{s\pi}{2}\right)$$