Condution of complex function's one to one

Question

Can you prove the following theorem? Let $G\subset \mathbb{C}$ is a convex region and $f:G\rightarrow \mathbb{C}$ analytic function. $\forall z\in G, Re(f'(z))>0\Rightarrow f$ is one-to-one function over $G$.

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 $D$.