Let $G$ be a convex area and $f:G \rightarrow \mathbb C$ be a holomorphic function (with continious derivative). Prove: If $|f'(z)-1|\lt 1,$ then f is injective on $G$

Could anyone give me a hint ? I dont know how to tackle this problem

  • 2
    $\begingroup$ FTC + $\operatorname{Re} f'(z) > 0$ $\endgroup$ – Daniel Fischer May 14 '17 at 19:20

Suppose $f(z_1) = f(z_2)$ for $z_1 \not = z_2$. Then look at the line $\gamma$ connecting $z_1$ to $z_2$. Since $G$ is convex, this line lies in $G$ and so $f$ is holomorphic here. Then $0 = f(z_2)-f(z_1) = \int_{\gamma} f'(z)dz = \int_0^1 f'(z_1+(z_2-z_1)t)\cdot(z_2-z_1)dt$

$\implies \int_0^1 f'(z_1+(z_2-z_1)t)dt = 0 \implies \int_0^1 f'(z_1+(z_2-z_1)t)-1 dt = -1$ but $|\int_0^1 f'(z_1+(z_2-z_1)t)-1dt| \le \int_0^1 |f'(z_1+(z_2-z_1)t)-1|dt < \int_0^1 1dt = 1$, a contradiction.

  • $\begingroup$ Still upvoted even though I specifically asked for a hint and not a solution. However, I appreciate your answer. $\endgroup$ – XPenguen May 14 '17 at 19:32
  • $\begingroup$ Also, it's much easier to do it in the manner Daniel Fischer suggested. $\endgroup$ – zhw. May 14 '17 at 21:41
  • $\begingroup$ Sorry about that $\endgroup$ – mathworker21 May 15 '17 at 3:57

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