Let $f$ be a non-constant holomorphic(analytic) function in the unit disc$\{|z|<1\}$ such that $f(0)=1$
then it is necessary that
$(1)$ there are infinitely many points $z$ in the unit disc such that $|f(z)|=1 $
$(2)\ f$ is bounded
$(3)$ there are at most finitely many points in the unit disc such that $|f(z)|=1$
$(4)$ $f$ is rational function.
I tried the if (2)is true then and it is given that $f$ analytic then it will be constant.Am I right? How to disprove of prove other option($\textbf{ Here answer is unique}$).please help me.thanks in advance.
Best Answer
Hints:
Note that (1),(3) are the negation of one another. This implies that for any give $f$, exactly one of them is the case (but, perhaps not that different examples cannot admit different cases).
Restate the context: $f$ is analytic on $\Bbb D$ and $1\in S^1\cap\mathrm{Rng}(f)$. We're asked whether $|S^1\cap\mathrm{Rng}(f)|$ is finite or infinite.
Added I think it's safe to elaborate on my earlier hints. Truth be told, though, I haven't noticed that the answer should best rely on the maximum modulus principle. In particular, I based my remarks on the open mapping theorem (which the maximum modulus principle can be viewed as a special case of). Indeed, if the range of $f$ contains $1$, it also contains some neighborhood thereof, which intersects a segment of the unit circle $S^1$. Since all these points must have distinct preimages, $f$ maps to $S^1$ on infinitely (uncountably) many points of $\Bbb D$.