MHB Is $f$ a one-to-one function if $\text{Re}(f') > 0$ on a convex set $X$?

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Here's this week's problem!

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Problem. Let $f$ be holomorphic function on a convex set $X \subset \Bbb C$ such that $\text{Re}(f') > 0$ on $X$. Show that $f$ is one-to-one.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

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No one answered this week's problem. You can find my solution below.

Spoiler

Since $X$ is convex, if $z_1$ and $z_2$ are distinct points of $X$, then \[\frac{f(z_2) - f(z_1)}{z_2 - z_1} = \int_{[z_1,z_2]} f'(z)\, \frac{dz}{z_2 - z_1} = \int_0^1 f'((1 - t)z_1 + tz_2)\, dt.\] Taking real parts and using the condition $\text{Re}(f') > 0$, we find that
\[\text{Re}\left(\frac{f(z_2) - f(z_1)}{z_2 - z_1}\right) > 0.\] Thus $f(z_1) \neq f(z_2)$.