suppose $|a|<1$, show that $\frac{z-a}{1-\overline{a}z}$ is a mobius transformation that sends $B(0,1)$ to itself.

Question

Suppose $|a|<1$, show that $f(x) = \frac{z-a}{1-\overline{a}z}$ is a mobius transformation that sends $B(0,1)$ to itself.

To make such a mobius transformation i tried to send 3 points on the edge to 3 points of the edge. so filling $i,1,-1$ in $f(x)$ we should get on the edges of the unit ball. But i don't seem to know how to calculate these exactly:

$$f(1) = \frac{1-a}{1-\overline{a}1}$$

$$f(-1) = \frac{-1-a}{1+\overline{a}}$$

$$f(1) = \frac{i-a}{1-\overline{a}i}$$

I don't seem to get how i could write these formula's in such a way that i get into the edges of the circle.

Anyone can help me?

Kees

Tim Raczkowski · Accepted Answer · 2015-03-13T17:19:54.987

4

If $|z|=1$, then $\overline z=1/z$. So,

$$\left|{z-a\over 1-\overline az}\right|=\left|{z-a\over1-\overline{a/z}}\right|=\left|{\overline z(z-a)\over\overline z-\overline a}\right|=|\overline z|\left|{z-a\over\overline{z-a}}\right|=|\overline z|=1.$$

edited Mar 13 '15 at 17:19

answered Mar 13 '15 at 17:13

Tim Raczkowski

20,046

why is $\frac{z-a}{\overline{z}-\overline{a}} = 1$ ? – Kees Til Mar 13 '15 at 17:16
Because $\overline z-\overline a=\overline{z-a}$, and conjugates have the same modulus. – Tim Raczkowski Mar 13 '15 at 17:18
hmmmm, i'm not convinced maybe i do not understand it wel. Because suppose $z-a = w$. then we can write w as $re^{i\phi}$. how is this term: $\frac{re^{i\phi}}{re^{-i\phi}} = e^{2i\phi} = 1$? – Kees Til Mar 13 '15 at 17:24
if you choose $\phi = \frac{\pi}{4}$ this fails right? – Kees Til Mar 13 '15 at 17:26
True, ${re^i\phi\over re^{-i\phi}}\ne 1$ in general, but $\left|{re^{i\phi}\over re^{-i\phi}}\right|=\frac rr=1$. – Tim Raczkowski Mar 13 '15 at 17:28
o yeah i forget the absolute value brackets, thanks man! – Kees Til Mar 13 '15 at 17:36
Your welcome. Glad I could help. – Tim Raczkowski Mar 13 '15 at 17:39

score 1 · Answer 2 · answered Mar 13 '15 at 17:05

1

hint: prove that : $\left|\dfrac{z-a}{1-\bar{a}z}\right| < 1$ by using $z = x+iy, a = m+in$, and use the well-known CS inequality: $(a^2+b^2)(c^2+d^2) \geq (ac+bd)^2$.

answered Mar 13 '15 at 17:05

DeepSea

77,651

great didn't think about that! – Kees Til Mar 13 '15 at 17:09
i tried this but i don't get the form like that, did you multiplie both sides with $1+\overline{a}z$? – Kees Til Mar 13 '15 at 17:38

suppose $|a|<1$, show that $\frac{z-a}{1-\overline{a}z}$ is a mobius transformation that sends $B(0,1)$ to itself.

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