how to prove that a given function is univalent

Question

I have to prove that following function is univalent

$f(z) = z^2 +3z +1, ~|z|<1$ in complex plane.

What I tried is:

Let $f(z_1) = f(z_2)$ $\Rightarrow$ ${z_1}^2 +3z_1 +1= {z_2}^2 +3z_2 +1$ $\Rightarrow$ $(z_1 - z_2)(z_1 +z_2 +3) = 0$ How to prove that $(z_1 +z_2 +3)\neq0$ in the given domain.

I tried using the fact that $Re(z)\leq |Re(z)|\leq |z|$ but no success.

Any hint or suggestion would be helpful. Thanks for your help

Answer 1

Note that: $$|z|< 1\implies -1<\Re(z)<1.$$ Now, if $z_1+z_2+3= 0$ then

$$0=\Re(0)=\Re(z_1+z_2+3)=\Re(z_1)+\Re(z_2)+3\in(1,5),$$ which is impossible. Thus, $z_1+z_2+3\ne 0.$