How can we prove that $| f ( z ) | = \left| e ^ { i \theta } \frac { z - \alpha } { 1 - \bar { \alpha } z } \right| \leq 1$ for $| \alpha | < 1$ and $| z | \leq 1$
We have $\left| \frac { z - \alpha } { 1 - \bar{\alpha} z } \right| = | z | \left| \frac { 1 - \alpha / z } { 1 - \bar { \alpha } z } \right|$,
Can you show that $\left| \frac { 1 - \alpha / z } { 1 - \bar { \alpha } z } \right| = 1$ ?
Prove the inequality when $|z|=1$ and then use Maximum Modulus Principle.
The inequality reduces to $(1-|z|^{2})(1-|\alpha|^{2}) \geq 0$ when $|z|=1$.