Complex function’s injection(solve it please......) Let $D=\{z| |z-z_0|&lt;1\}$
If $f$ is analytic function on $D$ and $|f’(z)-f’(z_0)|&lt;|f’(z_0)|$,
Then $f$ is injective

First, I’m sorry
This is my first time to use this page
So, I want to solve this question.
I heard this question is very hard
But I believe you can solve it.
I have been studying Complex analsis
I spent time to solve this question
I couldn’t prove this question..
Please answer me, How to solve it?
Assume wlog $z_0=0$ (change variables $z-z_0=w$) and rotate and scale s.t $f'(0)=1$ (this doesn't change injectivity or lack of it); then the condition becomes $|f'(z)-1|&lt;1$ for all $|z|&lt;1$  and that implies $\Re f'(z) &gt;0$  (since $|a-1|&lt;1$ implies by squaring $|a|^2-2\Re a &lt;0$).

But now the unit disc is convex so the easy Noshiro Warschavski theorem (discussed here in other posts and obtained by integration on segments) applies so $f$ is injective
From the statement it follows that, on the disk, $f’$ doesn’t vanish. 

But we can actually state more: define on the disk $g(z)=f’(z_0)(z-z_1)$; then $|(f-f(z_1))’-g’| &lt; |g’|$. Since $(f-f(z_1))(z_1)=g(z_1)=0$, and since $g’$ is constant, we can integrate and it follows $|(f-f(z_1))-g| &lt; |g|$.

So by Rouché, in any disk of $D$, $f-f(z_1)$ and $g$ have the same number of roots. As a consequence, $\{f=f(z_1)\}=\{z_1\}$. 

Edit: in answer to your comment, the result is the following. Let $U$ be a convex open subset of $\mathbb{C}$ and let $f,g$ be holomorphic functions on $U$ such that $g$ is affine, $|f’| &lt; |g’|$ and for some $p$ in $U$, $f(p)=g(p)=0$. Then, everywhere in $U$ except at $p$, $|f| &lt; |g|$. 

Proof: let $z \in U$, $z \neq p$, $$|f(z)|=\left|(z-p)\int_0^1{f’((1-t)p+tz)\,dt}\right|=|z-p|\int_0^1{|f’((1-t)p+tz)|\,dt} &lt; |z-p|\int_0^1{|g’((1-t)p+tz)/,dt}=|z-p|\left|\int_0^1{g’((1-t)p+tz)\,dt}\right|=|g(z)|.$$