Inequality with absolution value for complex number

Question

How to show that inequality:

$|1-\bar{\alpha} z| \ge |z-\alpha|$

$z$ and $\alpha$ are complex number, $\alpha$ is constans and $|z|<1$, $| \alpha| < 1$

I can proof that by using substition $z=x+yi$ and $\alpha=t+ki$ but I'm looking for solving without that.

Thank you for every hint and help.

Answer 1

Consider $|1 - \bar{\alpha}z|^2 -|z - \alpha|^2$:

\begin{align}&|1 - \bar{\alpha}z|^2 - |z - \alpha|^2\\ &= (1 - 2\operatorname{Re}(\alpha \bar{z}) + |\alpha|^2 |z|^2) - (|z|^2 - 2\operatorname{Re}(\alpha \bar{z}) + |\alpha|^2)\\ &= 1 - |z|^2 - (1 - |z|^2)|\alpha|^2\\ &= (1 - |\alpha|^2)(1 - |z|^2)\\ & > 0. \end{align}

Thus $|1 - \bar{\alpha}z| > |z - \alpha|$.

Answer 2

Note that

$$|1-\bar{\alpha} z|^2=(1-\bar{\alpha} z)(1-\alpha \bar{z})=1-\bar{\alpha}z-\alpha\bar{z}+|\alpha z|^2$$ and

$$|z-\alpha|^2=(z-\alpha)(\bar{z}-\bar{\alpha})=|z|^2+|\alpha|^2-\bar{\alpha}z-\alpha\bar{z}.$$ So, we need to show that

$$1+|\alpha z|^2>|z|^2+|\alpha|^2.$$ But previous inequality can be written as

$$1-|\alpha|^2> |z|^2(1-|\alpha|^2),$$ and, since $|\alpha|<1,$ it is

$$1> |z|^2,$$ which holds by assumption.