But is there another solution for $i^i=x^x$?

Question

So I was scrolling through Youtube again when I found this video by Blackpenredpen which defined all real and imaginary solutions for$$x^x=y^y$$which got me interested, because it got me asking$$\text{Is there another solution for }i^i=x^x\text{?}$$

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$$\text{Here is my attempt at finding another solution to }i^i=x^x$$

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$$i^i=x^x$$$$i\log_xi=x$$$$\log_xi=-ix\text{ because }\frac{x}{i}=\frac{x}{\sqrt{-1}}=\frac{x\sqrt{-1}}{-1}=-ix$$$$x^{log_xi}=x^{-ix}$$$$i=\frac{1}{x^{ix}}$$$$ix^{ix}=1$$$$ix^{ix}-2=e^{i\pi}\text{ or }ix^{ix}=e^{i\pi}+2$$$$x^{ix}=-ie^{i\pi}-2i$$$$ix\ln(x)=\ln(-ie^{i\pi}-2i)$$$$x\ln(x)=-i\ln(-ie^{i\pi}-2i)$$$$x\ln(x)=-i\ln(-i(-1)-2i)$$$$x\ln(x)=-i\ln(i-2i)$$$$x\ln(x)=-i\ln(-i)$$$$x^x=(-i)^{-i}$$$$\therefore i^i=(-i)^{-i}$$My question

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Is it correct that $i^i$ does in fact equal $(-i)^{-i}$, or is there a flaw in my proof?

Answer 1

This is equivalent to solving $z^z=e^{-\pi/2}$

$z\ln z=-\pi/2$.

$r(\cos \theta +i\sin\theta)(\ln r+i\theta)=-\pi/2$

$r[(\cos \theta \ln r -\theta \sin \theta)+i(\sin\theta \ln r+\theta\cos \theta)]=-\pi/2$

$\sin\theta \ln r+\theta\cos\theta=0\implies r=e^{(-\theta \cot\theta)}$

$e^{-\theta \cot \theta}(-\theta \cos^2 \theta \csc \theta-\theta\sin \theta)=-\pi/2$

$e^{-\theta \cot \theta}=(\pi /2) \sin \theta / \theta$

$-\theta \cos \theta/\sin \theta = \ln(\pi /2)+\ln (\sin \theta /\theta) $

I don't think that last line can be solved analytically.

Answer 2

Throughout your calculations, you've used $i = \frac1{-i}$. Raising it to $i$th power: $$\boxed{i^i = \frac{1}{(-i)^i} = (-i)^{-i}}$$

Answer 3

You have to be careful with branches: logarithms and non-integer powers of complex numbers are multivalued functions. Thus $\log(i) = i \pi/2 + 2 \pi i n$ so $$ i^i = \exp(i \log(i)) = \exp\left(-\frac{\pi}{2} - 2\pi n\right)$$ where $n$ is an arbitrary integer. The principal branch is $n=0$. As for $-i$, we have $\log(-i) = -i \pi/2 + 2 \pi i n$ so $$(-i)^(-i) = \exp(-i \log(-i)) = \exp\left(-\pi/2 + 2 \pi n\right)$$ Again, $n=0$ is the principal branch, so the principal branches of $i^i$ and $(-i)^{-i}$ are the same.

Solutions of $x^x = c$ are $x = \log(c)/W(\log(c))$, where $W$ is any branch of the Lambert W function. Taking $\log(c) = -\pi/2$, the $0$ branch of $W$ gives you $i$ and the $-1$ branch gives you $-i$. There are also infinitely many other branches.