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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!

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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.

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  • $\begingroup$ Why $R$ is at most $R = |z_1 - z_0|$? $\endgroup$ – mathmaniac Sep 30 '18 at 11:09
  • $\begingroup$ Because otherwise the region would include the point $-a/2$, but the function is not injective in any neighborhood of that point. $\endgroup$ – Hugo Sep 30 '18 at 11:11
  • $\begingroup$ Why the function is not injective in neighborhood of point $z_! = -\frac{a}{2}$? $\endgroup$ – mathmaniac Sep 30 '18 at 11:14
  • $\begingroup$ 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. $\endgroup$ – Hugo Sep 30 '18 at 11:18
  • $\begingroup$ Ok, thank you very much for your answers! $\endgroup$ – mathmaniac Sep 30 '18 at 11:19
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With $w=z-z_0$ we can analysis $f(w)=w^2+Aw+B$ for univalency in $|w|<R$, 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 $R<\dfrac12|2z_0+a|$.

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