When does $i^x=x$

Question

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.

Answer 1

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.

Answer 2

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 $i^x=x$.