Given two non-zero complex numbers $z$ and $w$ such that $zw$ doesn't equal $-1$. Prove if $\overline{z}=z^{-1}$ and $\overline{w}=w^{-1}$, then $\frac{(z+w)}{(1+zw)}$ is real.
Having trouble simplifying the expression.
I know the denominator would always be real since I'll multiple by conjugate, I don't know about numerator.
On
To prove that $\frac{(z+w)}{(1+zw)}$ is real we have to show that $\frac{(z+w)}{(1+zw)}=\overline{\left(\frac{(z+w)}{(1+zw)}\right)}=\frac{(\bar z+\bar w)}{(1+\bar z \bar w)}$.
In other words, we have: $$\frac{(\bar z+\bar w)}{(1+\bar z \bar w)}=\frac{(\frac{1}{a+bi}+\frac{1}{c+di})}{(1+\frac{1}{a+bi} \frac{1}{c+di})}=\frac{\frac{(a+c)+i(b+d)}{(a+bi)(c+di)}}{\frac{(ac-bd)+i(ad+bc)+1}{(a+bi)(c+di)}}=\frac{(a+c)+i(b+d)}{(ac-bd)+i(ad+bc)+1}$$ On the other hand, we have: $$\frac{(z+w)}{(1+zw)}=\frac{(a+c)+i(d+b)}{1+(ac-bd)+i(adbc)}$$ We have shown that $\frac{(z+w)}{(1+zw)}=\frac{(\bar z+\bar w)}{(1+\bar z \bar w)}$, so $\frac{(z+w)}{(1+zw)}$ is real.
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Q $$ \begin{aligned} \bar{z}=\frac{1}{z}, \bar{\omega}=\frac{1}{\omega}, \Rightarrow|z| & =1=|\omega \mid \\ \because \frac{z+\omega}{1+z \omega}-\left(\frac{\bar{z}+\bar{\omega}}{1+\bar{z} \bar{\omega}}\right) & =\frac{z+\omega}{1+z \omega}-\left(\frac{1 / z+1 / \omega}{1+1 / z \omega}\right) \\ & =\frac{z+\omega}{1+z \omega}-\frac{z \omega}{z \omega}\left(\frac{\frac{1}{z}+\frac{1}{\omega}}{1+\frac{1}{z \omega}}\right) \\ & =\frac{z+\omega}{1+z \omega}-\left(\frac{\omega+z}{z \omega+1}\right)=0 . \end{aligned} $$
Thus; $\Rightarrow \frac{z+\omega}{1+z \omega}=\frac{\bar{z}+\bar{\omega}}{1+\bar{z} \bar{\omega}}=\left(\overline{\frac{z+\omega}{1+z \omega}}\right), \Rightarrow \frac{z+\omega}{1+z \omega}$ is purely neal
$$F=\frac{z+w}{1+zw} \implies F=\frac{\frac{1}{w}+\frac{1}{z}}{\frac{1}{z}\frac{1}{w}+1}=\frac{\bar w+ \bar z}{\bar z \bar w+1}= \bar F$$ So $F$ is real.