Complex analysis, Ian Stewart Exercise 4.7.5: Proving $\sqrt{z}$ is continuous on $\mathbb{C}\setminus\{x\leq0\}$

Question

This is exercise 4.7.5 in Ian Stewart's "Complex Analysis (The Hitch Hiker’s Guide to the Plane)":

Let $C_{\pi} =\{z\in\mathbb{C}:z\neq x\in\mathbb{R},x\leq0\}$ be the 'cut plane' with the negative real axis removed. Define $r : C_{\pi} \to \mathbb{C}$ by $(r(z))^2 = z$ and $\operatorname{Re} r(z) > 0$. Prove that $r$ is continuous on $C_{\pi}$.

May I ask how can I prove it without using theorems like Inverse function theorem from multivariable calculus?

Answer 1

If $z\in\Bbb C\setminus(-\infty,0]$, then$$r(z)=\frac{\sqrt{\operatorname{Re}(z)+|z|}}{\sqrt2}+\frac{\operatorname{Im}(z)}{\sqrt2\sqrt{\operatorname{Re}(z)+|z|}}i,\label{a}\tag1$$since:

1. $\displaystyle\left(\frac{\sqrt{\operatorname{Re}(z)+|z|}}{\sqrt2}+\frac{\operatorname{Im}(z)}{\sqrt2\sqrt{\operatorname{Re}(z)+|z|}}i\right)^2=z$;
2. $\displaystyle\operatorname{Re}\left(\frac{\sqrt{\operatorname{Re}(z)+|z|}}{\sqrt2}+\frac{\operatorname{Im}(z)}{\sqrt2\sqrt{\operatorname{Re}(z)+|z|}}i\right)=\frac{\sqrt{\operatorname{Re}(z)+|z|}}{\sqrt2}>0$.

It is clear that the RHS of \eqref{a} is continuous.