There is only one $z\in \Bbb C$ such that $z^2=w$ and $Re(z)>0$

Question

Let $w\in\Bbb C$, show that unless $w\in \Bbb R^-$, there exists only one $z\in \Bbb C$ such that $z^2=w$ and $Re(z)>0$.

This question is related to this other question, but this is a modification of the problem to have a solution.

Answer 1

Consider the complex polynomial equation in the variable $z\colon z^2-w=0.$ It has two distinct solutions for every non-zero complex constant $w$. Take for $w$ a number that is not negative real. If the solutions are denoted $a,b$, check that $(a/b)^2= 1$. Or $a=\pm b$. Clearly one of $a$ or $b$ will have positive number as its real part.