If $\operatorname{Re}f^\prime > 0$ on a convex domain, then $f$ is one-to-one.

Question

Let $f(z)$ be analytic on a convex region $D \subset \mathbb{C}$. If $\mathrm{Re}f'(z)>0,\forall z\in D$, then show that $f(z)$ is a one-to-one function, that is, if $z_1\ne z_2,$ then $f(z_1)\ne f(z_2)$.

Answer 1

Let $f \colon D \to \mathbb{C}$ be analytic on a convex subset $D$ with $\operatorname{Re}f^\prime(z) > 0 ~~ \forall z \in D$.

Let $z_1 \neq z_2$ be two distinct points in $D$. By the mean-value theorem for holomorphic functions, there exists some point $z_0$ on the line segment between $z_1$ and $z_2$ satisfying

$$\operatorname{Re}f^\prime(z_0) = \operatorname{Re} \left( \dfrac{f(z_2) - f(z_1)}{z_2 - z_1} \right) > 0.$$

The assumption that $D$ is convex guarantees that the line segment from $z_1$ to $z_2$ is contained in $D$ — in fact, this is the definition of a convex subset.

And since the real part of the derivative is positive (by hypothesis), we cannot have $f(z_1) = f(z_2)$, since then we would have $\operatorname{Re}f^\prime(z_0) = 0$. Thus $f(z_1) \neq f(z_2)$, so $f$ is injective (one-to-one) on $D$.