I was working on the following nested square root problem:
Let $a \in \mathbb R ^+$, what is the value of: $$\sqrt{a \sqrt{a \sqrt{a}}}...$$
I concluded that the answer is $a$ and then I thought about generalizing this to the complex numbers. The question became:
Let $z \in \mathbb C$, what is the value of: $$\sqrt{z \sqrt{z \sqrt{z}}}...$$
I did the following:
Let $x \in \mathbb C$ with $x =\sqrt{z \sqrt{z \sqrt{z}}}... $. Then: $$\sqrt{z \sqrt{z \sqrt{z}}}... = \sqrt{z} \sqrt{\underbrace{\sqrt{z \sqrt{z \sqrt{z}}}...}_x}$$
So we concude that $x = \sqrt{z} \sqrt{x}$ or: $$x^2 = |z| |x| \ \ (1)$$.
Let $x = r e^{i \phi}$ and $ z = p e^{i \alpha}$. Then, eq. (1) becomes:
$$r^2e^{i 2 \phi} = pr$$
$$r = 0\vee r = p e^{- i 2 \phi}$$
We know that $r \neq 0$ so we have that $r = p e^{- i 2 \phi}$. because $x = r e^{i \phi}$ if we substitute this we get:
$$x = p e^{- i \phi}$$
So this concludes that:
$$\left| \sqrt{z \sqrt{z \sqrt{z}}}... \right| = |z|$$
My question is: I was able to find the absolute value of the solution, now to fully have a solution for this problem I need the argument of it. How can find the argument of the solution?
Over the entire complex field, it's not possible to consistently (continuously) define $\sqrt .$ everywhere. It's the same problem as defining a logarithm : you have to cut somewhere.
So your question won't have a definite answer. My guess is : for every phase $\varphi$, there is a definition of $\sqrt .$ such that your answer is $|z|e^{i\varphi}$.
The idea is that the series $\sum \frac {x_i} 2$ for $(x_i) \in \{0,1\}^{\mathbb N}$ can converge to anything in $[0,1]$ so by defining the cut to realise the right series $(x_i)$ we can add any phase we want.