Complex analysis

[Sk_Anonymous]

Let f be analytic on {|z| <1}.
Suppose |f ' (z)-f ' (0)|< |f ' (0)| for all |z|<1.
Prove f is injective.


Thanks for your help.
karl P.S. "Io,speriamo che me la cavo"

[Luca.Lussardi]

By contradiction we assume that there exists $z_1$ and $z_2$ with $|z_i| <1$ and with $z_1 \ne z_2$, $f(z_1)=f(z_2)$. Let $\gamma(t)$ be the segment joining $z_1$ with $z_2$ parametrized by arc length. Then
$0=f(z_1)-f(z_2)=\int_{\gamma} f'(z)dz=f'(0)(z_2-z_1)+\int_{\gamma}(f'(z)-f'(0))dz$. Thus
$\int_{\gamma}(f'(z)-f'(0))dz=-f'(0)(z_2-z_1)$. But
$\int_{\gamma}(f'(z)-f'(0))dz=\int_0^{|z_2-z_1|}(f'(\gamma(t))-f'(0))\gamma'(t)dt$
and then
$|f'(0)| |z_2-z_1|=|\int_{\gamma}(f'(z)-f'(0))dz| \le \int_0^{|z_2-z_1|}|f'(\gamma(t))-f'(0)| |\gamma'(t)|dt < |f'(0)| |z_2-z_1|$ which is a contradiction.

[Camillo]

Karl, welcome to the English section of our Forum

Nice exercise, not easy..