Exponential relationship issue

Question

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.

Answer 1

Use basic index laws.

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

Answer 2

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 $i$

Answer 3

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 $1/4$.