Estimate for domain of the analytic function with nonzero derivative to be injective Let $f$ be analytic on the open unit disc whose range is also contained in the open unit disc. Let $f(0)=0$ and $f'(0)=a \neq 0$
Then I know that there is disc of positive radius $\mid z \mid < p$ such that $f$ is injective in the open disc. However, How do I estimate the value of $p$ in terms of $a$? This is a new problem and I cannot find a way through. Could anyone please help me?
 A: This is a Theorem of Landau and has as answer (obviously by Schwarz $|a| \le 1$):
$p=\frac{1}{|a|}(1-\sqrt{(1-|a|^2)}= \frac{|a|}{1+\sqrt{(1-|a|^2)}}$
The proof is not that hard but it takes a while; if we let $p=p(a)$ the radius above and $w=\frac{p-z}{1-pz}$, $f(z)=\frac{w^2-p^2}{p^2w^2-1}$, satisfies $f(0)=0, f'(0)=|a|, |f(z)| <1, f'(p)=0$, so it shows that $p$ is indeed maximal (so nothing bigger can work - a rotation by the argument of $a$ clearly gives an example with $f'(0)=a$)
To actually show that $p$ works is more involved.
Edit later - I looked through my library and could find this only in volume 2 of Goodman's Univalent Functions (chapter 12, pg 52, Theorem 13) and that is a harder to find book then Duren (Univalent Functions - main reference for the title subject or say Garnett (Bounded Analytic Functions - main reference for the title subject), so I will sketch the proof presented there.
First, notice that actually the example that shows that $p=p(|a|)$ is an upper bound can be re-written simply as:
$f(z)=e^{i\theta}\frac{z(|a|-z)}{1-|a|z}=az+z^2f_1(z)$, 
where $a=e^{i\theta}|a|$ and now it is really straightforward to check the required properties, most notably $f'(p)=0$ as the rest easily follow from $f$ being a Blaschke product of order 2- the complicated form above arises from the Dieudonne Theorem about the radius of $q$-valence for bounded analytic functions on the unit disc (with non-zero derivative at the origin) which generalizes Landau's theorem when $q=1$
Second, let's now have $g(0)=0, g'(0)=|a|, |g| <1$ (by a rotation we can clearly get that from $g_1'(0)=a$ without modifying the univalence radius) and call $R$ the univalence radius of $g$, so we need to prove $R \ge p$. We also can assume $|a| <1$ so $p <|a|$ as otherwise by Schwarz, $g$ is just $az$ 
We can assume $R \le p < |a|$ as otherwise we are done, so then there are points $c_{1,2}$ on the circle of radius $R$, s.t $g(c_{1,2})=b$ (where we allow $c_1=c_2$ as double root, so then $g'(c_1)=0$), while $g$ is univalent in the disc of radius $R$ centered at the origin (though this is not needed, just $g$ having a double point on the $R$ circle).
We consider $h(z)=\frac{g(z)}{z}$. By Schwarz $|h| <1$ in the unit disc, while $h(0)=|a|$, hence since $|c_1|=R <|a| <1$, we have $|h(c_1)| \ge \frac{|h(0)|-R}{1-R|h(0)|}$, or 
$|b| \ge R\frac{|a|-R}{1-R|a|} >0$
But now let $h_1(z)=\frac{g(z)-b}{1-\bar b g(z)}\frac{1-\bar c_1 z}{c_1-z}\frac{1-\bar c_2 z}{c_2-z}$ 
It is easy to see that if $h_2(z)=\frac{g(z)-b}{1-\bar b g(z)}$ and $h_3(z)=\frac{c_1-z}{1-\bar c_1 z}\frac{c_2-z}{1-\bar c_2 z}$, then $h_2(c_{1,2})=0, |h_2| <1$ on the unit disc, while $h_3$ is Blaschke, so it is $1$ in absolute value on the unit circle, so by maximum modulus $h_1(z)=\frac{h_2(z)}{h_3(z)}$ is analytic and of modulus at most $1$ in the unit disc ($h_2$ having a double zero at $b$ if $c_1=c_2$), which gives 
$1 \ge |h_1(0)|= \frac{|b|}{R^2}$
Putting the two inequalities together we get:
$R^2 \ge |b| \ge R\frac{|a|-R}{1-R|a|}$ or $|a|R^2-2R+|a| \le 0$ and this is precisely the condition that defines $p(a)$ as the root of the above between $0$ and $1$, with the other root greater than $1$ so indeed $p(a) \le R < 1$, and we are finally done.
