Show that $f(z)=(1+z)^2$ is univalent in open unit disk $|z|<1$

Question

univalent function

I have tried to show that the function $f(z)=(1+z)^2$ as univalent function in unit disk $|z|<1$
as steps I'm doing are

Let $f(z_1)=f(z_2)$ for any $z_1, z_2$ in the unit disk
then $(1+z_1)^2 = (1+z_2)^2$

$1+z_1^2+2z_1 = 1+z_2^2+2z_2$

$z_1^2+2z_1 = z_2^2+2z_2$

what to do next to show that
$z_1 = z_2$?

Answer 1

Let $f(z_1)=f(z_2)$, and $\mathbb U$ is the unit disk, then \begin{split} f(z_1)-f(z_2) & = (1+z_1)^2 - (1+z_2)^2 \\ & = (1+z_1 - 1 - z_2)(1+z_1+ 1+ z_2) \\ & = (z_1 - z_2)(2+z_1+ z_2) \quad (z_1, z_2 \in \mathbb U) \end{split} implies that $z_1 = z_2$, because if $z_1+ z_2 = -2$ then $2 = |z_1+z_2| < |z_1|+|z_2| < 2 $ which is a contradiction.