Showing that $b\bar{d}=a\bar{c}\hspace{3mm}\text{and} \hspace{3mm} a\bar{b}=c\bar{d}$ & $\frac{P(z)}{Q(z)}=\omega \frac{z-\alpha}{1-\bar{\alpha}z}$

Question

Let $P(z)=az-b$ and $Q(z)=cz-d$, where $a,b,c,d$ are nonzero complex numbers such that $bc\neq ad$.

Suppose that $|\frac{P(z)}{Q(z)}|=1$ whenever $|z|=1$. Show that

$$b\bar{d}=a\bar{c}\hspace{3mm}\text{and} \hspace{3mm} a\bar{b}=c\bar{d}$$

and deduce that

$$\frac{P(z)}{Q(z)}=\omega \frac{z-\alpha}{1-\bar{\alpha}z}$$

where $\alpha=\frac{b}{a}$ and $\omega$ is a complex number with modulus $1$.

$\textbf{My attempt so far:}$

Since $|\frac{P(z)}{Q(z)}|=1$ whenever $|z|=1$, then

$$|az-b|^2=|cz-d|^2 \rightarrow (az-b)\overline{(az-b)}=(cz-d)\overline{(cz-d)}$$

$$\rightarrow (az-b)(\bar{a}\bar{z}-\bar{b})=(cz-d)(\bar{c}\bar{z}-\bar{d})$$

$$\rightarrow a\bar{a}|z|^2-a\bar{b}z-\bar{a}b\bar{z}+b\bar{b}=c\bar{c}|z|^2-c\bar{d}z-\bar{c}d\bar{z}+d\bar{d}$$

$$\rightarrow |a|^2 -(a\bar{b}-c\bar{d})z+|b|^2=|c|^2-(\bar{c}d-\bar{a}b)\bar{z}+|d|^2$$

I'm stuck at this point. Am I on the right track, or is my method of approach wrong on a whole?

Answer 1

$\quad |a|^2 -(a\bar{b}-c\bar{d})z+|b|^2=|c|^2-(\bar{c}d-\bar{a}b)\bar{z}+|d|^2$

Almost done at this point, now multiply with $\,z\ne 0\,$ and use that $\,z \bar z = |z|^2 = 1\,$ one more time:

$$(a\bar{b}-c\bar{d})z^2- \left(|a|^2+|b|^2-|c|^2-|d|^2\right)z +(\bar{a}b-\bar{c}d) = 0$$

The latter quadratic has every $\,z\,$ on the unit circle as a root, thus infinitely many roots, and therefore must be the zero polynomial. So, in particular, $\,a\bar{b}-c\bar{d} = 0 \iff a \bar b = c \bar d\,$.

For the other equality, note that if $\,f(z) = \dfrac{az-b}{cz-d}\,$ preserves the unit circle, then so does its inverse $\,f^{-1}(z)=\dfrac{dz-b}{cz-a}\,$. Applying the argument above to $\,f^{-1}\,$ gives $\,d\bar b = c \bar a \iff a \bar c = b \bar d\,$.