I was messing with the identity e^{i\pi}=-1 and I got that i = \sqrt{e^{\pi\sqrt{e^{\pi\sqrt\ldots}}}} and on. I plugged it in to a calculator and it was infinite. It grew very fast. Does that make i solvable?

**Answer**

I’m assuming to reach that conclusion, you took the square root of both sides and got:

\sqrt{e^{i\pi}}=i

which is correct. Then, on the left hand side, you substituted for i using this equality to get

\sqrt{e^{\pi\sqrt{e^{i\pi}}}}=i

which is still okay – and then you tried doing this infinitely often, and lo and behold, it broke!

The main issue is that, suppose we let

f(x)=\sqrt{e^{\pi x}}.

What’s you’ve shown amount to the fact that f has what is called a *fixed-point* at i – that is

f(i)=i.

And this implies things like

f(f(i))=i

f(f(f(i)))=i

or, where we let f^n(i) be f applied to i repeatedly n times, we get

f^n(i)=i.

So far so good. However, if we wanted to make sense of an expression which is basically an infinite nest of the function f with no starting value, it would be that, no matter what x we started with, the sequence (x,f(x),f^2(x),f^3(x),\ldots) – called the orbit of x under f – would always converge to i – and as you see, this is not the case.

A simple example of the difference between “has a fixed point” and “converges to something” would be if we took

f(x)=2x

which has a fixed point at x=0. If we wished to think of the expression 2\cdot (2\cdot (2\cdots)))

we would hope that f would have that (x,f(x),f^2(x),\ldots) converged for any x – but it only converges for x=0 – otherwise, the sequence has f^n(x)=2^nx which diverges. Your function acts similarly, but it is harder to visualize, as it takes place on the complex plane. So, in fact, you cannot “compute” i using this method – the series will simply diverge. However, for any *finite* number of applications, your identity is correct.

**Attribution***Source : Link , Question Author : user2888499 , Answer Author : user26486*