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Can someone point out the flaw here?

$$e^{-3\pi i/4} = e^{5\pi i/4}$$

So raising to $\frac{1}{2}$, we should get

$$e^{-3\pi i/8} = e^{5\pi i/8}$$

but this is false.

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How many values can the complex square root have? How do you choose consistently? –  Mark Bennet Oct 26 '12 at 6:45
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Why should $e^{\theta i}=e^{\phi i}$ imply $e^{\theta i/2}=e^{\phi i/2}$? A obvious example is $e^{\pi i}=e^{-\pi i}$ but $e^{\pi i/2} \ne e^{-\pi i/2}$. –  wj32 Oct 26 '12 at 6:47
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2 Answers 2

Paraphrase using $e^0=1$ and $e^{\pi i}=-1$. We can write $$ e^{-3\pi i/4}\;1^2=e^{-3\pi i/4}\;(-1)^2 $$ Raising to the $\frac12$ power yields $$ e^{-3\pi i/8}\;1=e^{-3\pi i/8}\;(-1) $$ The problem is that without proper restrictions (e.g. branch cuts), the square root is not well-defined on $\mathbb{C}$.

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The problem is that $(e^x)^y=e^{xy}$ does not hold with complex numbers as it does with real numbers. This can change what the principal value, which is what has happened in your example. You should read a bit about principal logarithms and branch cuts.

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