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So I was thinking, $\exp(i\theta) = \exp( i\theta\cdot2\pi\cdot\frac{1}{2\pi})$, we can rearrange it, so that:

\begin{align} & \exp\left( i\theta\cdot2\pi\cdot\frac{1}{2\pi}\right)=\exp\left(2\pi i\cdot\frac{\theta}{2\pi}\right)=\exp(2\pi i)^{\theta/(2\pi)} \\[8pt] = {} & (\exp(\pi i)^2)^{\theta/(2\pi)}=((-1)^2)^{\theta/(2\pi)}=1^{\theta/(2\pi)}=1 \end{align}

Can someone tell me what went wrong?

Michael Hardy
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walterudoing
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    You need to be careful when taking complex roots or you can do something similar and get $(\sqrt{-1})^2=1$ by say, the following: $\sqrt{-1}\cdot\sqrt{-1}=\sqrt{(-1)^2} = \sqrt{1}=1.$ What went wrong is $x^\frac{a}{b}\neq (x^{a})^\frac{1}{b}$ in general with complex roots. – Maxim Gilula Feb 05 '15 at 20:39

1 Answers1

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$$ e^{(ab)} \overset{\text{?}}{=} (e^a)^b $$

The equality above is true and $a$ and $b$ are real. Your example shows that it doesn't always work when they are complex but not real.

And that holds for other bases than $e$ as well. Consider $1^{1/3}$. Certainly that should mean a cube root of $1$, since $1^{1/3}\cdot1^{1/3}\cdot1^{1/3}$ should be $1^{1/3+1/3+1/3}=1^1$. But there are three different complex cube roots of $1$. Which one is it?

Michael Hardy
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