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Question:
Show that $$\color{Blue}{e^{-\pi/2}=i^{i}}$$

My answer:
First establish that $\qquad\qquad\quad e^{-i\pi/2}=\cos(-\pi/2)+i \sin(-\pi/2)=-i$
then $\qquad\qquad\qquad\qquad \mathrm{Log}(e^{-i \pi/2})=\mathrm{Log}(-i)$
so $\qquad\qquad\qquad\qquad\quad -i \mathrm{Log}(e^{ -\pi/2})=\mathrm{Log}(-i)$
$\qquad\qquad\qquad\qquad\qquad\quad \mathrm{Log}(e^{ -\pi/2})=-i^{-1}\mathrm{Log}(-i)$

Therefore the RHS should equal to $\mathrm{Log}(i^{i})$, but how?

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    $1/i=i/i^2=-i$. – 2017-02-12
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    Please be aware that logarithms are multi-valued functions when we have complex values so $i^i$ can be equated to multiple values. In general you know that if $z=re^{i\theta}=re^{i(\theta+2\pi n)},\ n\in\mathbb{N}$ so $\log z=\log r + i\theta=\log r+i(\theta+2\pi n)$. – 2017-02-12
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    I rolled back because the capitalized Logs are likely intentional, specifying the principal branch of the logarithm. The capitalization was in the original. – 2017-02-12
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    Actually you are using Euler's Notation in a wrong way. See my answer – 2017-02-12
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    Is it a coincidence that the owner of the duplicated OP and this OP have the same username? – 2017-02-12

2 Answers 2

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There is a shorter way to answer the problem. You have $\log i = \ln|i| + i\arg(i) = i \dfrac{\pi}{2}\mathbb{Z}$. It has lot of branches. If $\operatorname{Log}$ denotes the principal branch of the logarithm, then its imaginary part at $i$ is simply $\dfrac{\pi}{2}$. Now $$ i^i = e^{i\operatorname{Log} i} = e^{i(i\frac{\pi}{2})} = e^{-\frac{\pi}{2}}. $$

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    Don't like the answer? Comment, please! – 2017-02-12
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    I don't downvote this but I can answer your comment See here http://math.stackexchange.com/questions/2085268/does-i4-equal-1 – 2017-02-12
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HINT: Use Euler's Notation $$e^{i\theta}=\cos\theta+i\sin\theta$$ Use Polar form that $$i=\cos{\frac{\pi}{2}+i\sin\frac{\pi}{2}}=e^{i\frac{\pi}{2}}$$ Now taking $i$ as power both the sides $$i^i=\left(e^{i\frac{\pi}{2}}\right)^i$$ $$e^{i^2\frac{\pi}{2}}=i^i$$ $$e^\frac{-\pi}{2}=i^i$$

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    I think downvoters must tell the reason of downvote – 2017-02-12
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    There is no such requirement on stackexchange. My guess would be that your answer doesn't address the OP's issue of getting $-i^i\log(-i)$ to equal $\log i^i$ but its just my thought. – 2017-02-12
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    But I am helping the OP to solve the **original question** because of which he has this problem – 2017-02-12
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    As I said just my thought. Perhaps you could provide insight to the OP as to how they technique is inappropriate. Some mention of the multi-values nature of complex logarithms could help. – 2017-02-12
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    (I didn't downvote it.) Looks OK to me except that part of taking $i$ as power on both sides. – 2017-02-12
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    But whats wrong in that – 2017-02-12
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    How do you conclude that $\left(e^{i\frac\pi2}\right)^i=e^{i^2\frac\pi2}$? Is the principle $\left(e^a\right)^b=e^{ab}$ for complex numbers $a$ and $b$? That would have limited validity. There are different possible values depending on chosen branches, not addressed here. E.g., in the question there appears to be a definition based on the principal branch of the logarithm, and it isn't clear how that is used here. – 2017-02-12
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    The original version of this answer only stated Euler's identity, which was already used in the question. When I looked later and saw the question had not been cleared up in this answer, but instead an alternative approach that glossed over the definition needed to make the conclusion, removal of downvote seemed unwarranted. – 2017-02-12
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    Actually OP himself used this concept here that $$\log e^{-i \frac{\pi}{2}}=-i\log{e^{\frac{\pi}{2}}}$$ – 2017-02-12