Proof that in any set A such that A contains a circumference centered at zero you can't find a continuous square root function.

Question

The following is an exercise in my textbook on complex analysis:

Proof that in any set A such that A contains a circumference centered at zero you can't find a continuous square root function.

I'm trying to use to following definitions and theorems:

A function $h:\Omega \to \mathbb{C}$ is a square root in $\Omega$ if $h(z)^2=z$ $\forall z\in\Omega$.

Theorem 1: Let $\Omega \subseteq \mathbb{C}-\{0\}$ a non-empty and open set. The following are equivalent:

1. There are continuous arguments in $\Omega$.
2. There are continuous logarithms in $\Omega$.
3. There are holomorphic logarithms in $\Omega$.
4. The function $z \to \frac{1}{z}$ has a primitive function (antiderivative) in $\Omega$.

Theorem 2: Let $\Omega \subseteq \mathbb{C}-\{0\}$ non-empty containing a circumference centered at zero, then A hasn't got continuous arguments.

I'm not sure I need to use these but I believe some of ideas could be reused.