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.

  • 1
    $\begingroup$ I had a suggestion just now, which I deleted - I think it is easier than that, because $u/v$ is a real number iff $u$ and $v$ lie on the same line through the origin. Namely, choose $z$ such that $a$ and $b$ are on a line through $z$. $\endgroup$ – peter a g Dec 1 '15 at 4:21
  • $\begingroup$ @peterag: Thank you. It helps me lot to find my own solution. What do you think about the case $$f(z)=\dfrac{(z-a)(z-c)}{(z-b)(z-d)}\,\,\, ?$$ $\endgroup$ – Bumblebee Dec 1 '15 at 7:34

$$\mathbf a=p+qi$$ $$\mathbf b=u+vi$$ $$\mathbf z=x+yi$$ 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, $$\frac{y-q}{x-p}=\frac{y-v}{x-u}=m$$ $$f(\mathbf z)=\frac{x-p+(y-q)i}{x-u+(y-v)i}=\frac{x-p}{x-u}\cdot\frac{1+\frac{y-q}{x-p}i}{1+\frac{y-v}{x-u}i}=\frac{x-p}{x-u}\cdot\frac{1+mi}{1+mi}=\frac{x-p}{x-u}\in\mathbb R$$


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