Which branch of the square root allows $(z^2)^{1/2} = z$

Question

I am aware that $(z^2)^{1/2} = z$ does not hold for every branch of the function $f(z) = z^{1/2}$.

For example if we do not take a branch and consider $z=-1$, we get $$f((-1)^2) = \exp(\frac{1}{2}Log((-1)^2)) = exp(\frac{1}{2}Log(1) = \exp(0) = 1$$.

ie if we square and then square root -1, we do not get back to -1.

Is it possible to pick a branch of $f(z) = z^{1/2}$ such that when we do $f(z^2)$, we get back to the same root that we started with?

Thanks

Answer 1

You need to choose the branch that gives you $\sqrt 1=-1$, which is not the principal one, but the one defined as $$ F(z)=exp(\frac 1{2}[\ln |z|+i arg\,z+2\pi i]). $$ Then, you end up with $$ [(-1)^2]^{1/2}=exp(\pi i)=-1. $$ In general, if you choose for $z$ an argument in $[0,2\pi)$, your relation will hold true with the principal branch if $arg\,z\in [0,\pi)$ and with the second branch if $arg\, z\in[\pi,2\pi)$.

Answer 2

If the implicit universal quantifier is meant, i.e., "$(z^{2})^{1/2} = z$ for all complex $z$" then no, it's impossible: The squaring map $g(z) = z^{2}$ is not injective. For example, $g(1) = 1 = g(-1)$; no branch of square root can send $1$ back to both $1$ and $-1$.