Let $f:\mathbb{C}\to \mathbb{C}$, $f(z)=z^2$ and $B = \{z \in \mathbb{C},Re(z)\leq0\}$. Show that $f^{-1}(B)$ its a closed set

Let $f:\mathbb{C}\to \mathbb{C}$, $f(z)=z^2$ and $B = \{z \in \mathbb{C},Re(z)\leq0\}$.

Show that $f^{-1}(B)$ its a closed set.

This is my attempt:

Let $w \in B, w =a+bi, a\leq 0$

Putting in the polar form:

$|w|\cdot \cos(\theta) = a \implies \cos(\theta)=\frac{a}{\sqrt{a^2+b^2}}\leq 0$ $\cos(\theta)\leq 0 \implies \frac{\pi}{2}\leq\theta\leq\frac{3\pi}{2}$

I think that the interval of values to the argument of w its relevant to the question. But I dont know how to proceed from this point.

I tried to part of the definition of a inverse function: $$ f: A \to B $$ $$ f^{-1}(X) = \{x\in A: f(x) \in X\} $$ But i didnt get anywhere as well

Solution 1:

If you write in coordinates $$z^2=(x+i y)^2 = (x^2 - y^2) + 2 i x y$$

so $$\{z \ | \mathcal{Re}(z^2)\le 0 \} = \{ x + i y \ | x^2 - y^2 \le 0\}$$ a full cone.