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\begin{align*} 1&>e^{-4\pi^2} \\ &= e^{(2\pi i)(2\pi i)} \\ &= (e^{2\pi i})^{2\pi i} \tag{$\ast$} \\ &= 1^{2\pi i}\\ &= 1 \end{align*}

The problem with this fake proof is that if we considered the principal branch of logarithm, $\operatorname{Log}$:

$$\large(e^{2\pi i})^{2\pi i}=e^{\operatorname{Log}(e^{2\pi i})2 \pi i}= e^{(\log(1)+i \cdot 0)2 \pi i}=1$$

So, the second equality would be false and the fake proof uses the fact that the logarithm is multi-valued for the exponentiation. Is this idea correct?

Gary
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J P
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  • look at here and here – wasu Apr 15 '24 at 20:09
  • $\operatorname{Log}(\mathrm{e}^{2\pi \mathrm{i}})=\log(|\mathrm{e}^{2\pi \mathrm{i}}|)+\mathrm{i}\arg(\mathrm{e}^{2\pi \mathrm{i}})=0+\mathrm{i},2\pi=2\pi \mathrm{i}$. – Gary Apr 16 '24 at 01:54
  • @Gary, using the principal branch of logarithm, shouldn't the principal argument be $0$ ($-\pi<0<\pi$)? – J P Apr 16 '24 at 07:22
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    I was using the multivalued logarithm. What you say in your question is correct. – Gary Apr 16 '24 at 07:24

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