# 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.
• Why $R$ is at most $R = |z_1 - z_0|$? – mathmaniac Sep 30 '18 at 11:09
• Because otherwise the region would include the point $-a/2$, but the function is not injective in any neighborhood of that point. – Hugo Sep 30 '18 at 11:11
• Why the function is not injective in neighborhood of point $z_! = -\frac{a}{2}$? – mathmaniac Sep 30 '18 at 11:14
• Take $w_1 = -a/2 + w$ and $w_2 = -a/2-w$, and verify that $f(w_1)=f(w_2)$ for any choice of $w$, by substituting in the equation. – Hugo Sep 30 '18 at 11:18
With $$w=z-z_0$$ we can analysis $$f(w)=w^2+Aw+B$$ for univalency in $$|w|, then if $$w_1\neq w_2$$ we see $$f(w_1)-f(w_2)=(w_1-w_2)(w_1+w_2+A)\neq0$$ if $$(w_1+w_2+A)\neq0$$ or $$|w_1+w_2|<|A|$$, from $$|w_1+w_2|<|w_1|+|w_2|<2R$$ we should have $$2R<|A|$$ or $$R<\dfrac12|2z_0+a|$$.