# 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.

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There must be an error. The quantity on the left is equal to $e^{-\pi x/2}$ exactly; the extra $i$ is not correct. – Michael Grant Apr 26 '13 at 12:17
I don't think the relation is true. The expression on the left is $(-i)^{-ix}$. There is no universal convention for raising $-i$ to a non-integer power, which is what you are trying to do. – Stefan Smith Apr 26 '13 at 23:31

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 $1/4$.

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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 $i$

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yeah that is what i get but the factor i is important, mm il have another think – user73685 Apr 26 '13 at 12:07
@user73685 can you give some context? It may be very important – Jorge Apr 26 '13 at 12:09

Use basic index laws.

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

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This isn't true in general for complex $a$ and $b$ -- not even as a multivalued expression. – 6005 Apr 26 '13 at 13:30
It wasn't a complete answer, and that was for good reason. – Glen O Apr 26 '13 at 13:42