Univalent function with parameter. Let $a, b, z_0 \in \mathbb{C}$. Find the highest value R, at which the function: $f(z) = z^2 + az + b$ is univalent in the disk: $|z - z_0| < R$ I use the definition of univalent function: If $z_1 \neq z_2 \Rightarrow f(z_1) \neq f(z_2)$. Using this, I got $a = -(z_1 + z_2)$. Then if i am not mistaken anywhere I got: $z_0 = 0$ or $z_0 = z_1 + z_2 = -a$. Thus I have: $|z| < R$ or $|z + a| < R$. How it's help me? Thank you very much in advance!

You are not on the right track. The derivative of $f(z)$ is $f'(z) = 2z+a$, which is zero at the point $z_1 = -a/2$. Hence the highest value of $R$ is at most $R = |z_1-z_0|$. On the other hand it is easy to see that the function is injective on the region $$ U = \\{ z : |z - z_0| < |z_1 - z_0| \\}, $$ as by the change of variables $w = z + a/2$ your function is $f(w) = w^2 + c$, and the behaviuor of $w \mapsto w^2$ around zero is well known and easy tu study.