This seems a quite easy concept, but I feel like I am missing something. The function $f(z) = \sqrt{(z-a)(z-b)}$, with $a, b \in \mathbb{C}$, is a multi-valued function, and as such, it's true that $$ \sqrt{(z-a)(z-b)} = \sqrt{(z-a)}\sqrt{(z-b)} $$
meaning that all values in LHS are also in RHS. Now, my textbook is introducing branch cuts and branch points theory, and in order to study $f(z)$, the function is rewritten as $$ f(z) = \sqrt{|z-a||z-b|}\exp{\frac{\theta_1 + \theta_2}{2}} $$
with $\theta_1, \theta_2$ being the relative angles of z with respect to $a$ and $b$ respectively. It is then shown that $a$ and $b$ are branch points, by calculating $f(z)$ along a closed loop around them. This is clear. What confuses me is the next passage.
To study $f(z)$ at infinity, the usual substitution $z \rightarrow 1/z$ is made, thus calculating $$ f(1/z) = \sqrt{\frac{(1-za)(1-zb)}{z^2}} = \frac{\sqrt{(1-za)(1-zb)}}{z} \sim z $$ concluding that infinity is not a branch point. But we used $\sqrt{z^2} = z$, which is just not true; in fact $\sqrt{z^2} = \pm z$. I understand that we would conclude the same about infinity not being a branch point, but the fact that it is proved like this leaves me with many confusing doubts.
