Let $$f(z)=\dfrac{z-a}{z-b},\,\,\,\,\,\,z\not=b\not=a$$ be a complex valued rational function.
How can I show that,
if $|a|,|b|\lt1,$ then there is a complex number $z_0$ satisfying $|z_0|=1$ and $f(z_0)\in\mathbb{R}$ ?
I have tried in many ways, but on success. Basically I tried to show that there is a unimodular complex number such that $$\dfrac{a-b}{z-b}=\dfrac{\bar a-\bar b}{\bar z-\bar b}.$$ I could make a quadratic equation by using the fact that $\bar z=\dfrac{1}{z},\,\forall z\in\partial\Bbb{D}.$ Unfortunately I could not solve this question using that. So, I would like to see different (and somewhat general) approach.
Also, I would like to know that what happen if (both or atleast one) $a,b\not\in\Bbb{D}.$
Any comment or hint will be welcome. Thank you.
$$\mathbf a=p+qi$$ $$\mathbf b=u+vi$$ $$\mathbf z=x+yi$$ We can always find $\mathbf z$ so that $\mathbf a$, $\mathbf b$ and $\mathbf z$ are on a same line (actually whether $|\mathbf z|=1$ or not, or $|\mathbf a|,|\mathbf b|<1$ or not).
Then, $$\frac{y-q}{x-p}=\frac{y-v}{x-u}=m$$ $$f(\mathbf z)=\frac{x-p+(y-q)i}{x-u+(y-v)i}=\frac{x-p}{x-u}\cdot\frac{1+\frac{y-q}{x-p}i}{1+\frac{y-v}{x-u}i}=\frac{x-p}{x-u}\cdot\frac{1+mi}{1+mi}=\frac{x-p}{x-u}\in\mathbb R$$