Is it possible to construct a function $g$ which is analytic on an open set containing the closed unit disc such that $g(z)=(\Im (z))^{2023}$, for all $z$ in the boundary of the unit disc?
I'm pretty much stuck on this problem and not quite sure how to proceed. Any help/hint will be very useful. Thanks in advance.
The idea here is that any (nonconstant) analytic function that is real on the unit circle, must have a singularity (which we can make a pole) either inside the unit circle or on the unit circle itself, so the answer to the question is negative; here, for example, $f(z)=((z-1/z)/2i)^{2023}$ works with precisely a pole at the origin; the proof is very easy by the open property of analytic functions.
Let $f$ analytic in the unit disc, continuous and real on the unit circle; then $f(\mathbb D)=U$ a bounded domain in the plane and $f(\partial \mathbb D)=K$ a compact, connected real set, so a closed interval. But (because $f$ is an open map) $\partial U \subset f(\partial \mathbb D)=K$ which means that $U$ is unbounded and that is a contradiction.
If we allow poles inside the unit disc, or we allow $f$ on the unit circle to take values in $\mathbb R \cup \infty$ (and be continuous in the extended sense) then we get such examples easily as noted.