Showing that $\left|\frac{z-a}{1-\bar{a}z}\right|=1$

Suppose $a$ and $z$ are in $\mathbb{C}$ with $|z|=1$ and $|a|<1$. I want to show $$\left|\frac{z-a}{1-\bar{a}z}\right|=1$$ but I'm stuck on dealing with $a$. Is this just another complex number of the form $a+bi$?

How do I show this is true? Thanks in advance.

Solution 1:

Multiply the left side by $\dfrac{1}{\bar{z}}$. That does not change the norm, since $z$ has norm $1$. The result is $$\left|\frac{z-a}{\bar{z}-\bar az\bar{z}}\right|.$$ But $z\bar{z}=1$, and therefore $$\left|\frac{z-a}{\bar{z}-\bar az\bar{z}}\right|=\left|\frac{z-a}{\bar{z}-\bar a}\right|.$$ But $\bar{z}-\bar{a}$ is the conjugate of $z-a$, so the ratio has norm $1$.

Solution 2:

$$z\cdot\bar z=1\implies1-\bar a\cdot z=z\cdot(\bar z-\bar a)\implies|1-\bar a\cdot z|=|z|\cdot|\bar z-\bar a|=1\cdot|z-a|$$

Solution 3:

as $a\neq z$ then$$\left|\frac{z-a}{1-\bar{a}z}\right|=\frac{\left|z-a\right|}{\vert \overline{1-\bar{a}z}\vert}=\frac{\left|z-a\right|}{\left|1-a\bar{z}\right|}=\frac{\left|z-a\right|}{\left|z\bar{z}-a\bar{z}\right|}=\frac{\left|z-a\right|}{\left|\bar{z}\right| \left|z-a\right|}=1$$