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.
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} &amp;= e^{-ix \log (e^{-i \pi/2 })} \\
&amp;= e^{-ix (-i\pi/2 + 2\pi i k) } &amp;&amp; (k \text{ an integer; } k = 0 \text{ for principle log})\\
&amp;= e^{-x\pi/2 + 2\pi x k} \\
&amp;= 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} &amp;= i e^{-x\pi/2} \\
\iff e^{2\pi k x} &amp;= i \\
\iff e^{2\pi k x} &amp;= e^{i \pi / 2} \\
\iff 2 \pi k x &amp;\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 $0 \le r &lt; 1$ real.
\begin{align*}
2 \pi k (i(m + r)) &amp;\equiv i \pi / 2 \pmod {2\pi i} \\
\iff 2 \pi i m k + 2 \pi i k r &amp;\equiv i \pi / 2 \pmod {2\pi i} \\
\iff mk + kr &amp;\equiv 1 / 4 \pmod {1} \\
\iff kr &amp;\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 $1/4$.
Use basic index laws.

$$
(e^a)^b = e^{ab}
$$
What do you get when you do this?