Show that $f(z)=2z+z^2$ with $|z|<1$ is a one-to-one function.

By using $z=x+iy$, I get
$$f(z)=x^2+2x-y^2+i(2y+2xy)$$

So to prove that this function is one-to-one, suppose $f(z\_1)=f(z\_2)$ where $z\_1=x\_1+iy\_1,z\_2=x\_2+iy\_2$, by comparing the real part and imaginary part,
$$x\_1^2+2x\_1-y\_1^2=x\_2^2+2x\_2-y\_2^2$$
$$2y\_1+2x\_1y\_1=2y\_2+2x\_2y\_2$$.

So how can I reach the result $x\_1=x\_2,y\_1=y\_2$? And how can I relate the information $|z|<1$ in proving?