Exponential relationship issue I read this relation and I am not sure why this is true, is it I can't see why it would be?
$$(e^{ -i\pi/2})^{ -ix}\approx ie^{-\pi x/2} $$ I get that $e^{i\pi/2}=-i$, but I can't see why this relation would be true. 
 A: You have to be careful working with complex exponentials because of problems with multivalued logarithms.  The exponential $a^b$ for $a,b \in \mathbb{C}$  is usually defined in terms of the multivalued logarithm as $e^{b \log a}$.
As a multivalued expression, we have
\begin{align*}
(e^{-i \pi/2 })^{-ix} &= e^{-ix \log (e^{-i \pi/2 })} \\
&= e^{-ix (-i\pi/2 + 2\pi i k) } && (k \text{ an integer; } k = 0 \text{ for principle log})\\
&= e^{-x\pi/2 + 2\pi x k} \\
&= e^{2\pi k x}e^{-x\pi/2}
\end{align*}
Therefore, for your statement to be true (in the sense that for one branch it works) we would have to have
\begin{align*}
e^{2\pi k x}e^{-x\pi/2} &= i e^{-x\pi/2} \\
\iff e^{2\pi k x} &= i \\
\iff e^{2\pi k x} &= e^{i \pi / 2} \\
\iff 2 \pi k x &\equiv i \pi / 2 \pmod {2\pi i}
\end{align*}
For this to be true, $x$ must be purely imaginary.  Let $x = i(m + r)$ where $m$ is an integer and $0 \le r < 1$ real.
\begin{align*}
2 \pi k (i(m + r)) &\equiv i \pi / 2 \pmod {2\pi i} \\
\iff 2 \pi i m k + 2 \pi i k r &\equiv i \pi / 2 \pmod {2\pi i} \\
\iff mk + kr &\equiv 1 / 4 \pmod {1} \\
\iff kr &\equiv 1 / 4 \pmod {1}
\end{align*}
In summary: the only time your statement could actually be sort of true is when $x$ is purely imaginary, and the fractional part of the imaginary part of $x$ (i.e. $r$ above) multiplied by some integer $k$ has a fractional part of $1/4$.
A: Given $z$, $\omega \in \mathbb{C}$, $z\neq 0$ 
$$z^{\omega}=e^{\omega Log(z)} $$
where $$Log(z)=log(|z|)+i(arg(z)) $$
Let $ z = e^{-i\frac{\pi}{2}},- $ $\omega=-ix$
Then 
$$(e^{-i\frac{\pi}{2}})^{-ix}=e^{-ix(Log(e^{-i\frac{\pi}{2}}))} $$
$$Log(e^{-i\frac{\pi}{2}})=log|1|+i(-\frac{\pi}{2})=-i\frac{\pi}{2} $$
so
$$(e^{-i\frac{\pi}{2}})^{-ix}=e^{-ix(-i\frac{\pi}{2})}=e^{-\frac{x\pi}{2}} $$
I'm sorry I don't get the factor $i$
A: Use basic index laws.
$$
(e^a)^b = e^{ab}
$$
What do you get when you do this?
