Complex numbers proof with modulus argument question

Question

I'm having trouble with this complex numbers proof.

Prove $|(z-a)/(\overline az-1)|=1$ if $a,z$ are any complex numbers, where $z\ne a$ and $|z|=1$.

I tried substituing $z,a$ for general Cartesian and polar forms, but I couldn't get past the algebra or required simplifications.

Answer 1

$$|(z-a)/(\overline az-1)|=1\iff|z-a|=|\overline az-1|$$ Now, since $|z|=1$, $$|\overline az-1|=|\overline a-1/z|=|\overline a-\overline z|=|a-z|=|z-a|$$

Answer 2

That's equivalent to $|z-a|=|\overline az-1|$. As $|z|=1$, $\overline z=z^{-1}$ and so $$|\overline az-1|=|z||\overline a -z^{-1}|=|\overline a -\overline z| =|\overline{a-z}|=|a-z|.$$

Answer 3

Another approach is using $z=e^{it}$ then $$\left|\dfrac{z-a}{\overline{a}z-1}\right|=\left|\dfrac{e^{it}-a}{\overline{ae^{-it}-1}}\right|=\dfrac{|e^{it}-a|}{|ae^{-it}-1|}=\dfrac{|e^{it}-a|}{|e^{-it}||a-e^{it}|}=1$$

Answer 4

Geometric proof:

• $|z - a| = |\bar z - \bar a|$ (conjugate is reflection along the horizontal axis)

• $|\bar z - \bar a| = |z\bar z - z\bar a|$ (product by $z$ with $|z| = 1$ is rotation)

• $|z\bar z - z\bar a| = |1 - z\bar a|$ (opposite rotations)

So $|1 - z\bar a| = |z - a|\ne 0$ because $z\ne a$ and we can divide by $|1 - z\bar a|$.