Is there anything special with complex fraction $\left|\frac{z-a}{1-\bar{a}{z}}\right|$?

Question

Is there anything special with the form:

$$\left|\frac{z-a}{1-\bar{a}{z}}\right|$$ ? With $a$ and $z$ are complex numbers.

In fact, I saw it in a problem:

If $|z| = 1$, prove that $|\frac{z-a}{1-\bar{a}{z}}| = 1$
If $|z| < 1$ and $|a| < 1$, prove that $|\frac{z-a}{1-\bar{a}{z}}| < 1$

I can easily prove the first one with expansion:

$$z=\cos\theta + i\sin \theta \\ a = m +in$$

But it will be terrible to use in the second one. What's more, I found this form a little special so maybe there is some clever trick without using expansion?

Thank you a lot!

This is very related (I won't go as far as saying it's a duplicate). — Najib Idrissi, Sep 06 '14 at 07:54
You can factor out $|z|$, which together with $\bar{z} = z^{-1}$ and one more observation gives the identity. Similar considerations also give an argument for (2). The ratio in your display equation is a Mobius transformation (in fact, all are given by products of that ratio and constants of modulus 1). Mobius transformations are precisely the linear fractional transformations that preserve the unit disc. Moreover, they're all invertible and so they form a group under composition that we can identify with $PGL(2, \mathbb{R})$. — Travis Willse, Sep 06 '14 at 07:59
@Travis: The transformation $z\mapsto -\frac{1}{z}$ is also a Moebius transform that does not preserve the unit disc. It is only the ones that you describe that do. — MichalisN, Sep 06 '14 at 08:11
@Michalis: Thanks for pointing this out, you're right, of course, I just managed to garble the statement during editing. — Travis Willse, Sep 06 '14 at 09:21
Does this answer your question? Can we characterize the Möbius transformations that maps the unit disk into itself? — tryst with freedom, Jun 28 '22 at 22:47

score 1 · Answer 1 · answered Apr 30 '15 at 02:36

1

We first prove that if $|z|=1$ then $|f(z)|=1$. For we have $$ |f(z)|=|\frac{z-a}{1-\bar az}|=|\frac 1{\bar z}\frac{z-a}{1-\bar az}|=|\frac {z-a}{\bar z-\bar a}|=1 $$ Now apply maximum principle to get $|f(z)|<1$ for $|z|<1$.

answered Apr 30 '15 at 02:36

k1.M

5,428

score 1 · Accepted Answer · answered Sep 06 '14 at 08:19

1

Hint: Use $z\overline{z}=|z|^2$ and compare $|z-a|^2$ to $|1-\overline{a}z|^2$.

Have a look at the comments to see why these transformations are special.

answered Sep 06 '14 at 08:19

MichalisN

5,402

Is there anything special with complex fraction $\left|\frac{z-a}{1-\bar{a}{z}}\right|$?

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