Let $x∈S^1$ and $a∈B$, where $S^1=\{z∈\mathbb{C}:|z|=1\}$ and $B$ denote the open disc in the complex field with centre $0$ and radius $1$. Let $M=x(\frac{z-a}{1-āz})$ for $z$ in $B$. Then show that $|M|<1$ if $|z|<1$. Here conjugate of $a$ is denoted by $ā$.
My attempt: Since $x∈S^1$ then $|x|=1$ and $a∈B$ then $|a|<1$ as $B$ is an open disc. Then $|M|=|x||\frac{z-a}{1-āz}|$. This implies $|M|=|\frac{z-a}{1-āz}|$. So, it is enough to show that $|\frac{z-a}{1-āz}|<1$ whenever $|z|<1$. But I failed to prove this. Please help me to solve this.
$$|M| \le \left|\frac{z-a}{1-\bar a z}\right| = \left|\frac{\bar z}{\bar z}\cdot \frac{z-a}{1-\bar a z}\right| = |\bar z| \left|\frac{z-a}{\bar z - \bar a}\right| = |\bar z| < 1$$