Let $f$ be holomorphic on $D'(0,1)=\{0<|z|<1\}$ and $f$ is continuous and real valued on $\{|z|=1\}.$ Show $f$ can be extended to $\mathbb{C}-\{0\}$ such that $f(z)= \overline{f(1/\overline{z})}, \forall z \neq 0.$
My attempt: $f=u+iv=u$ on $\{|z|=1\}.$ By Max/Min Principle for Harmonic functions, $v \equiv 0$ in $D^{\prime}(0,1) \ $ and so $f \equiv $ constant in $D'(0,1).$ By Identity Theorem, $f \equiv$ constant in $\mathbb{C} \ ?$ I suspect I have wrongly applied the Max/Min Principle because $f$ is not necessarily continuous on boundary of $D^{\prime}(0,1) \ ?$
I'm trying to make use the following theorem:
Let $\Omega$ be a region which is symmetric wrt to real axis and define $\Omega^+, \Omega^{-}$ and $L$ as intersection of $\Omega$ with upper half plane, lower half plane and real axis respectively. If $f$ is continuous on $\Omega^+ \cup L$ which is analytic on $\Omega^+$ and real on $L,$ then $f$ can be uniquely extended to holomorphic $F$ on all of $\Omega$ such that $F(z)=f(z), z \in \Omega^+ \cup L; F(z)= \ \overline{f(\overline{z})}, z \in \Omega^-.$
Could anyone advise please? Thank you.
Best Answer
Hint. Let $$ g(z)=f\left(\frac{z-i}{1-iz}\right). $$ Then $g$ takes real values on the real axis, and it is analytic in the upper half plane.
Use Schwarz Reflection for $g$, and then use the inverse transformation:
$$f(z)=g\left(\frac{z+i}{1+iz}\right).$$