An inequality of complex analysis?

Question

Show that for $0<\lvert \alpha\rvert <1$ and $\lvert z\rvert \leq r<1$ the inequality $$\left\lvert\frac{\alpha+\lvert\alpha\rvert z}{(1-\overline{\alpha}z)\alpha}\right\rvert \leq\frac{1+r}{1-r}$$ holds.

I consider this question as follows: let $$f_{\alpha}(z)=\frac{\alpha+\lvert\alpha\rvert z}{(1-\overline{\alpha}z)\alpha}$$ with $\lvert z\rvert \leq r<1$, and $f_{\alpha}$ is analytic in the closed disk $\lvert z\rvert \leq r$. So it gets its maximum modulus on the circle $\lvert z\rvert =r$.

Any further hint will welcome!

Answer 1

Following your idea: $f_{\alpha}$ gets the maximum value on the boundary $\vert z \vert =r$, then $$ \vert f_{\alpha}(z) \vert \leq \Big \vert \frac{\alpha+\vert \alpha \vert r}{(1-\bar{\alpha} z) \alpha}\Big \vert \leq \frac{\vert \alpha \vert +\vert \alpha \vert r}{\vert \alpha \vert- \vert \alpha \vert^2 r}\leq \frac{1+r}{1-\vert \alpha \vert r} \leq \frac{1+r}{1-r}.$$

Answer 2

I would maybe do the following, more algebraic, considerations: say $\lvert \alpha \rvert = \rho$ and $\lvert z \rvert = \sigma$. In particular, $\sigma \le r$. By the triangle inequality $$\lvert \alpha+\lvert \alpha \rvert z\rvert \le \rho(1+\sigma) \le \rho (1+r).$$ On the other hand, by $\lvert \lvert x \rvert-\lvert y \rvert \rvert \le \lvert x-y \rvert$:

$$\lvert (1-\overline{\alpha}z)\alpha \rvert = \lvert \alpha-\rho^2z\rvert \ge |\rho - \rho^2 \sigma\rvert = \rho(1-\rho \sigma) > \rho (1-r).$$ Summing up $$\frac{\lvert \alpha+\lvert \alpha \rvert z\rvert}{\lvert (1-\overline{\alpha}z)\alpha \rvert} \le \frac{\rho(1+r)}{\rho(1-r)} = \frac{1+r}{1-r}.$$

Answer 3

First note that $$ f_\alpha (z) = \left|\frac{\alpha+|\alpha|z}{(1-\overline{\alpha}z)\alpha}\right| = \left|\frac{\alpha+|\alpha|z}{\alpha-\left|\alpha\right|^2z}\right| = \left|\frac{u + z}{u - \left|\alpha\right|z}\right| $$ where $\displaystyle{u = \frac{\alpha}{\left|\alpha\right|}}$ is a unit vector, and $\left|\alpha\right|r < 1$. You can easily check that $\left|u + z\right|$ has a maximum $1+r$ at $z = ru$, and $\left|u - \left|\alpha\right|z\right|$ has a minimum $1-\left|\alpha\right|r$ at $z = ru$. So $f_\alpha(z)$ has a maximum at $\displaystyle{z = ru = \frac{r}{\left|\alpha\right|}\alpha}$ and then $$ f_\alpha(z) \le \frac{1+r}{1-\left|\alpha\right|r} \le \frac{1+r}{1-r}. $$