Showing that $\left|\frac{z-a}{1-\bar{a}z}\right|=1$

Question

Suppose $a$ and $z$ are in $\mathbb{C}$ with $|z|=1$ and $|a|<1$. I want to show $$\left|\frac{z-a}{1-\bar{a}z}\right|=1$$ but I'm stuck on dealing with $a$. Is this just another complex number of the form $a+bi$?

How do I show this is true? Thanks in advance.

Compute $\lvert 1 - \overline{a}z\rvert^2 - \lvert z-a\rvert^2$. — Daniel Fischer, Sep 18 '13 at 21:14
@DanielFischer I'm sure this is basic arithmetic but I don't know how to deal with $a$ - will you show me the first step? — User69127, Sep 18 '13 at 21:17
Treat it as a constant. It is one. An arbitrary constant, but a constant. — Daniel Fischer, Sep 18 '13 at 21:19

André Nicolas · Answer 1 · 2013-09-18T21:26:06.897

6

Multiply the left side by $\dfrac{1}{\bar{z}}$. That does not change the norm, since $z$ has norm $1$. The result is $$\left|\frac{z-a}{\bar{z}-\bar az\bar{z}}\right|.$$ But $z\bar{z}=1$, and therefore $$\left|\frac{z-a}{\bar{z}-\bar az\bar{z}}\right|=\left|\frac{z-a}{\bar{z}-\bar a}\right|.$$ But $\bar{z}-\bar{a}$ is the conjugate of $z-a$, so the ratio has norm $1$.

edited Sep 18 '13 at 21:26

answered Sep 18 '13 at 21:14

André Nicolas

507,029

I need help with $a$. Do i treat it as a constant? – User69127 Sep 18 '13 at 21:18
1

The number $a$ is a complex number. So is $z$. Both are, for the purpose of this problem, "constant." – André Nicolas Sep 18 '13 at 21:32
On that last part - does that mean that $\frac{x+iy}{x-iy}=1$? – whatwhatwhat Mar 03 '16 at 00:08
I don't know what you mean by $x+iy$, since that notation is not used in the answer. But it is not necessarily true that $\frac{z-a}{\bar{z}-\bar{a}}$ is $1$. The answer only asserts that its norm is equal to $1$. – André Nicolas Mar 03 '16 at 00:13

score 5 · Answer 2 · answered Sep 18 '13 at 21:28

as $a\neq z$ then$$\left|\frac{z-a}{1-\bar{a}z}\right|=\frac{\left|z-a\right|}{\vert \overline{1-\bar{a}z}\vert}=\frac{\left|z-a\right|}{\left|1-a\bar{z}\right|}=\frac{\left|z-a\right|}{\left|z\bar{z}-a\bar{z}\right|}=\frac{\left|z-a\right|}{\left|\bar{z}\right| \left|z-a\right|}=1$$

score 4 · Answer 3 · answered Sep 18 '13 at 21:19

4

$$z\cdot\bar z=1\implies1-\bar a\cdot z=z\cdot(\bar z-\bar a)\implies|1-\bar a\cdot z|=|z|\cdot|\bar z-\bar a|=1\cdot|z-a|$$

answered Sep 18 '13 at 21:19

Did

279,727

The hypothesis that $|a|\lt1$ is not needed. – Did Sep 18 '13 at 21:21
1

If $|a| = 1$ you need to interpret the quotient suitably though. – mrf Sep 18 '13 at 21:32

score 0 · Answer 4 · answered Dec 13 '13 at 01:42

0

since $\mid z \mid = 1$ the triangles $z,0,a$ and $1,0,a \overline{z}$ are congruent (by rotation about $0$), and the latter is anticongruent (by reflection in $Im(z)=0$) to $1,0,\overline{a}z$

answered Dec 13 '13 at 01:42

David Holden

18,040

Showing that $\left|\frac{z-a}{1-\bar{a}z}\right|=1$

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