When does $i^x=x$ Can someone please help me solve $i^x=x$?
So far I have:
$$i^x=x$$
$$\frac{\ln(x)}{\ln(i)}=x$$
$$e^{i\pi}=-1$$
$$e^{i\pi/2}=i$$
$$\frac{\ln(x)}{\frac{i\pi}{2}}=x$$
$$\ln(x)=\frac{i x \pi}{2}$$
$$2\ln(x)=i x \pi$$
$$e^{2ln(x)}=e^{ix\pi}$$
But from here, I don't know where to go (Also if I have any mistakes so far please let me know as well). Thank you very much.
 A: The Lambert W-function is a function $W(z)$ which solves $z=W(z)e^{W(z)}$. It is a multi-valued function.
In this case, you are trying to solve:
$$e^{x\pi i/2} = x$$
of:
$$\frac{-\pi i}{2}=\frac{-x\pi i}{2}e^{-x\pi i/2} $$
So 
$$x\frac{-\pi i}{2} = W(-\pi i/2)$$ or
$$x =\frac{2i}{\pi} W\left(\frac{-\pi i}{2}\right)$$
I don't think you can do better than this, and even then, this in only using one particular value for $i^x$ - you could also use $i^x=e^{5x\pi i/2}$.
This sort of problem is not even easy to solve when $i$ is replaced by a real number in $(0,1)$. For example:
$$\left(\frac{1}{2}\right)^x = x$$ clearly has a real solution in $(0,1)$, but no closed form without the $W$ function.
A: Thomas Andrews has already provided the general solution.  I thought that it might be instructive to see another approach when we assume that $x$ is real valued.  
Now, we know that $i=e^{i\pi(2\ell+1/2)}$ for all integers $\ell$.  Then, 
$$i^x=x\implies e^{i\pi(2\ell +1/2)x}=x \tag 1$$
Taking real and imaginary parts of $(1)$ yields
$$\cos (2\ell +1/2)\pi x=x \tag 2$$
and
$$\sin (2\ell +1/2)\pi x=0 \tag 3$$
From $(3)$ we see we must have $x=\frac{k}{2\ell +1/2}$, for any integer $k$, whereupon substituting back into $(2)$ reveals $$\cos k\pi =\frac{k}{2\ell +1/2}$$
which has no solutions.  Thus, there are no real solutions to $i^x=x$.
