# Complex analysis -pth roots

• Nov 26th 2011, 12:35 PM
tarheelborn
Complex analysis -pth roots
Let $f$ be a quadratic polynomial function of $z$ with two different roots $z_1$ and $z_2$. Given that a branch of $z$ of the square root of $f$ exists in a domain $D$, demonstrate that neither $z_1$ nor $z_2$ can belong to $D$. If $D$ had a double root, would the analogous statement be true?
• Nov 29th 2011, 05:12 PM
tarheelborn
Re: Complex analysis -pth roots
I know that $f(z)=a_0+a_1(z)+a_2(z^2)$ and I know that $g(z)^2=z$. So if $z$ has two different roots, then how could $g(z)^2 = z$ which implies that $\sqrt{z}$ and $z_1$ and $z_2$ are all in $D$. Is that where I need to go?