Show that a function satisfies $|f(z)|<1$ for all $z$ in the open left half plane.

The denominator of $f$ has two zeros $\frac{1\pm\sqrt{1-4a}}{2a}$, which we easily check to lie in the open right-half plane. So $f$ is holomorphic near $\overline{\mathbb{H}_L}$. Moreover,

• $f(z) \to 1$ as $z\to\infty \in \partial\mathbb{H}_L$, and
• $f(it) = \frac{1+it-at^2}{1-it-at^2}$ for $t\in\mathbb{R}$ satisfies $|f(it)| = 1$.

So $f$ satisfies $|f(z)| = 1$ on $\partial\mathbb{H}_L$. Since $f$ is a non-constant holomorphic function near $\overline{\mathbb{H}_L}$, the maximum modulus principle tells that $|f(z)| < 1$ on $\mathbb{H}_L$.

There is a direct proof.

Write $z=x+iy, x<0$.We have \begin{aligned} |f(z)|^2&=\frac{|1+x+iy+a(x^2-y^2)+2axyi|^2}{|1-x-iy+a(x^2-y^2)+2axyi|^2}\\&=\frac{(1+x+a(x^2-y^2))^2+(2axy+y)^2}{(1-x+a(x^2-y^2))^2+(2axy-y)^2}\\&=1+2x\frac{1+a(x^2-y^2)+2ay^2}{(1-x+a(x^2-y^2))^2+(2axy-y)^2}. \end{aligned} Finally, using $1+a(x^2-y^2)+2ay^2=ay^2+ax^2+1>0$ completes the proof.

Divide top and bottom by $z$, and notice that you obtain a composition of functions, one which obviously sends the left half plane to the unit disc and the other which stabilizes the left half plane. You should be able to fill in the details.

In more detail: we get the ratio $$\frac{(az+1/z) + 1}{(az+1/z)-1}$$

This is a composition of $z\mapsto az+1/z$ and $z\mapsto (z+1)/(z-1)$. The first map stabilizes the left half plane (because $a>0$) and the second maps the left half plane into the unit disc.