Let $|a| < 1$. Show that $1-|(z+a)/(1+\bar a z)|$ has the same sign as $1-|z|$

Question

Let $|a| < 1$. Show that $$1-\left|\frac{z+a}{1+\bar{a}z}\right|$$ has the same sign as $1-|z|$, where $a$ and $z$ are complex numbers.

As shown this sentence, I thought to split it in 2, when $1<|z|$ and $1>|z|$ and conclude that $$1<\left|\frac{z+a}{1+\bar{a}z}\right| \quad \text{and} \quad 1>\left|\frac{z+a}{1+\bar{a}z}\right|$$ respectively, but after doing a lot of bills I could not reach the wanted.

Hint: For $|a|<1$ $\frac{z+a}{1+\bar{a} z}$ is an automorphism of the unit disk. — user1337, Oct 10 '13 at 05:31
@AlecTeal it's an analytic map from the disk onto itself which has an analytic inverse. — user1337, Oct 10 '13 at 08:53
See for example http://math.stackexchange.com/questions/506058 I'm sure there are other near duplicates too — mrf, Oct 10 '13 at 09:24

score 1 · Answer 1 · answered Oct 10 '13 at 15:57

1

Use the formula $|a+b|^2=|a|^2+2\mathrm{Re}\,(a\bar b)+|b|^2$: $$|z+a|^2-|1+\bar a z|^2 = |z|^2+|a|^2-1-|a|^2|z|^2 = (|z|^2-1)\,(1-|a|^2)$$ and the conclusion follows.

answered Oct 10 '13 at 15:57

user100000

671

Let $|a| < 1$. Show that $1-|(z+a)/(1+\bar a z)|$ has the same sign as $1-|z|$

1 Answers1