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

Question

I've been stuck in problem stating that

Let $a>0$.Show that the function $$f(z)=\dfrac{1+z+az^{2}}{1-z+az^{2}}$$ satisfies $|f(z)|<1$ for all $z$ in the open left half plane.

Denote the open left half plane by $\mathbb{H}_{L}$.

Since this question deals with something on the $\mathbb{H}_{L}$, the first idea for me was to use conformal mapping so that we can sort of connect the open $\mathbb{H}_{L}$ to $\mathbb{D}$.

However, the thing here is that we have no idea where $f(z)$ maps $\mathbb{H}_{L}$ to, so the I compose in the following way.

Cleary the maps $$g:\mathbb{H}\longrightarrow\mathbb{H}_{L},\ \text{by}\ z\mapsto iz$$ and $$F:\mathbb{D}\longrightarrow\mathbb{H},\ \text{by}\ z\mapsto i\dfrac{1-z}{1+z}$$ are conformal.

Then, I compose $f\circ g\circ F:\mathbb{D}\longrightarrow \text{unknown}$.

I hoped that the composition map would finally look like an automorphism of $\mathbb{D}$.

However, it seems impossible to get rid of the $z^{2}$.

On the other hand, I tried to compose as $$F^{-1}\circ g^{-1}:\mathbb{H}_{L}\longrightarrow\mathbb{D}, $$which gives us a map $$z\mapsto\dfrac{1+z}{1-z}.$$

This one is closer to $f(z)$ except for there is no $az^{2}$ on either the numerator or the denominator.

Currently I have no idea about how to proceed.

Any idea or hints?

Thank you!

Answer 1

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.

Answer 2

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.

Answer 3

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 $\mathbb{H}_L$.