Proof using complex numbers

Prove that $\left|\dfrac{z-w}{1-\bar{z}w}\right| = 1$ where $\bar{z}$ is conjugate of $z$ and $\bar{z}w\ne 1$ if either $|z| = 1$ or $|w| = 1$.

I used $|c_1/c_2| = |c_1|/|c_2|$ and multiply out with $z = x + iy$ and $ = a+ib$ but I am getting stuck near finish.

Solution 1:

Important in complex analysis, Möbius transformation

1). it is easy to check that $\forall z_1, z_2$ $$|z_1-z_2|^2=|z_1|^2-2 \Re(z_1\bar{z}_2)+|z_2|^2$$

2) from 1), we have

$$|1-z_1\bar{z}_2|^2=1-2 \Re(z_1\bar{z}_2)+|z_1\bar{z}_2|^2$$

3). from 1) and 2), $\forall z_1, z_2$ $$|1-z_1\bar{z}_2|^2 -|z_1-z_2|^2 =(1-|z_1|^2)(1-|z_2|^2)$$

Solution 2:

Suppose $|z|=1$: then $$ \left|\dfrac{z-w}{1-\bar{z}w}\right|= \frac{1}{|z|}\left|\dfrac{z-w}{1-\bar{z}w}\right|= \left|\dfrac{z-w}{z(1-\bar{z}w)}\right|= \left|\dfrac{z-w}{z-z\bar{z}w}\right|= \left|\dfrac{z-w}{z-w}\right|=1 $$ If $|w|=1$, consider that $|1-\bar{z}w|=|1-\bar{w}z|$ because they are conjugate.

Solution 3:

Use the polar representation of the numbers $z$ and $w$, to let $|z|$ and $|w|$ appear.

If you expand $|z-w|^2=|1-\overline zw|^2$, you will see desired simplifications.