If $\Omega$ is convex and $|f'(z)-1|<1$ then $f$ is injective

The statement is: Let $$\Omega$$ be an open convex subset of $$\mathbb{C}$$, and $$f\in\mathcal{H}(\Omega)$$ with $$|f'(z)-1|<1$$, for all $$z\in\Omega$$. Prove that $$f$$ is injective.

I want to know if my proof is correct.

Suppose there exist $$\alpha,\beta\in\Omega$$ with $$\alpha\ne\beta$$ and $$f(\alpha)=f(\beta)$$. Because $$\Omega$$ is convex, we can integrate over the segment $$\alpha\beta$$, which is in $$\Omega$$. Then we have $$\left|\int_{\alpha}^{\beta} (f'(z)-1) \,dz\right| = \left|\left. f(z)-z\right]_{\alpha}^{\beta}\right|= |f(\alpha)-\alpha - f(\beta)+\beta| = |\beta -\alpha|.$$ By the other hand we have $$\left|\int_{\alpha}^{\beta}( f'(z)-1) \,dz\right|\leq \int_{\alpha}^{\beta}|f'(z)-1||dz| < \int_{\alpha}^{\beta} |dz| = |\beta - \alpha|$$ but this is not possible.

Is this correct? Thanks

• Fine........... Jul 2 '19 at 17:43
• @DavidC.Ullrich: Your edit makes my answer obsolete :) Jul 2 '19 at 17:44
• @MartinR Sorry - the answer hadn't appeared here when I made the edit. Jul 2 '19 at 17:46
• @MartinR I was upvoting your anwser when suddenly disapeared :(... thanks for the anwser Jul 2 '19 at 17:48