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Let's suppose that $|z|<1$ and $|w|=1$. Show that the modulus of $\displaystyle \frac{z-w}{1-\bar{z}w}$ is always $1$. Some hint.

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$|1-\bar{z}w| = |1-z\bar{w}|$ –  Simon Morris Sep 5 '12 at 14:25

4 Answers 4

up vote 6 down vote accepted

You can do a no thinking calculation. Let $\displaystyle u=\frac{z-w}{1-\bar{z}w}$. We calculate $u\bar{u}$. Note that $\displaystyle \bar{u}=\frac{\bar{z}-\bar{w}}{1-z\bar{w}}$. Thus $$u\bar{u}=\frac{z-w}{1-\bar{z}w}\cdot\frac{\bar{z}-\bar{w}}{1-z\bar{w}}=\frac{(z-w)(\bar{z}-\bar{w})}{(1-\bar{z}w)(1-z\bar{w})}.$$ Expand the denominator. We get $$(1-\bar{z}w)(1-z\bar{w})=1-\bar{z}w-z\bar{w}+z\bar{z}w\bar{w}=1-\bar{z}w-z\bar{w}+z\bar{z}.$$ Expand the numerator. We get the same thing.

Another way (or hindsight is $20$-$20$): In $\displaystyle \frac{z-w}{1-\bar{z}w}$, replace the $1$ by $w\bar{w}$. We get $\displaystyle \frac{1}{w}\cdot\frac{z-w}{\bar{w}-\bar{z}}$. But $\displaystyle \frac{1}{w}$ has norm $1$, as does $\displaystyle \frac{z-w}{\bar{w}-\bar{z}}$.

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Since $|w|=1$, $\bar w=\frac 1w$ hence $$\frac{z-w}{1-\bar z w}=\frac{z-w}{1-\bar z\frac 1{\bar w}}=\bar w\frac{z-w}{\bar w-\bar z},$$ which gives the result.

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Hint #1: If $|w|=1$ and $f$ is a complex-valued function, then $|f(z)|=|\overline{w}\cdot f(z)|$.

Hint #2: If $\alpha$ is a non-$0$ complex number, then $\left|\cfrac{\overline{\alpha}}{\alpha}\right|=1$.

Hint #3: $w\overline{w}=|w|^2$.

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for hint #1: I guess $|w|=1$? –  Fabian Sep 5 '12 at 15:25
/facepalm/ Yes, I meant "If $|w|=1$...," and it is now corrected. –  Cameron Buie Sep 5 '12 at 16:01

Hint $\ $ It is the special case $\rm\ a = 1-z\bar w,\,\ w\bar w = 1\: $ of the following

$$\rm b = -w \frac{a}{\bar a}\ \Rightarrow\ b\bar b\, =\, w\bar w \frac{a}{\bar a} \frac{\bar a}a\, =\, w\bar w$$

Remark $\ $ Thus it boils down to the the simple fact that $\rm\:a/\bar a\:$ has modulus $1,\:$ which holds more generally for norms of quadratic integers. A famous result of Hilbert (Theorem 90) asserts that the converse of this result is true in quadratic / cyclic extensions. This easily yields the parametrization of Pythagorean triples as a special case - see Olga Taussky's award-winning 1971 AMM paper Sums of Squares.

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