Complex analysis -pth roots

Let $\displaystyle f$ be a quadratic polynomial function of $\displaystyle z $ with two different roots $\displaystyle z_1$ and $\displaystyle z_2$. Given that a branch of $\displaystyle z$ of the square root of $\displaystyle f$ exists in a domain $\displaystyle D$, demonstrate that neither $\displaystyle z_1$ nor $\displaystyle z_2$ can belong to $\displaystyle D$. If $\displaystyle D$ had a double root, would the analogous statement be true?

Re: Complex analysis -pth roots

I know that $\displaystyle f(z)=a_0+a_1(z)+a_2(z^2)$ and I know that $\displaystyle g(z)^2=z$. So if $\displaystyle z$ has two different roots, then how could $\displaystyle g(z)^2 = z$ which implies that $\displaystyle \sqrt{z}$ and $\displaystyle z_1$ and $\displaystyle z_2$ are all in $\displaystyle D$. Is that where I need to go?