Existence of an analytic function on $\mathbb{D}$ where $\Re(f(z))=0$ for $z$ in a proper subarc of $\partial\mathbb{D}$.

Question

I would like to find a nonconstant analytic function $f:\mathbb{D}\to\mathbb{C}$, where $\mathbb{D}\subset\mathbb{C}$ is the unit disc and $f$ is continuous on $\bar{D}$ such that $\Re(f(z))=0$ for all $z$ in a proper subarc of $\partial\mathbb{D}$.

My first instinct is to make a branch cut outside of $\bar{\mathbb{D}}$ to obtain a branch of the square root function, thus taking $\mathbb{D}$ to the upper half circle, then apply the transformation $z\mapsto z+\frac{1}{z}$ to obtain (more or less) the bottom half plane. Then, I could rotate the bottom half plane to have its boundary "line up" with the real axis. Would this be sufficient? Please let me know if I am being unclear.

Answer 1

If $u$ is any real-valued $C^\alpha$ ($0 < \alpha < 1$) function, then the harmonic conjugate $v$ of the harmonic extension of $u$ to the disc is also $C^\alpha$.

In particular, any Hölder continuous function is the real part of a holomorphic function on $\mathbb{D}$, continuous on $\bar{\mathbb{D}}$. In particular you can make $u$ vanish on a large arc.

Unfortunately the same is not true assuming that $u$ is just continuous. There are continuous functions on the circle whose harmonic extension to the disc has an unbounded harmonic conjugate.

Answer 2

Let $\Gamma _0$ be a given proper subarc of $\partial\mathbb{D}$ and $\Gamma =\{z\mid z=e^{i\theta }, \pi\le \theta \le 2\pi \}$. Let $\varphi $ be a Moebius map $\varphi : \mathbb{D}\to \mathbb{D}$ with $\varphi (\Gamma _0)=\Gamma $.

We know that $$ \zeta =\left(\frac{1+w}{1-w}\right)^2$$ maps the upper half unit disk $\mathbb{D}^+=\{z\mid z\in \mathbb{D}, \Im z>0\}$ onto the upper half plane $\mathbb{H}^+$ and $$ z=\frac{\zeta -i}{\zeta +i}$$ maps $\mathbb{H}^+$ onto $\mathbb{D}$. Therefore $$z=\phi (w)=\frac{(1+w)^2-i(1-w)^2}{(1+w)^2+i(1-w)^2}$$ maps $\mathbb{D}^+$ onto $\mathbb{D}$. Note that $\phi $ maps $[-1,1]$ onto $\Gamma $. Thus $$ w=f(z)=\left(\phi ^{-1}\circ \varphi\right) (z)$$ maps $\mathbb{D}$ onto $\mathbb{D}^+$ with $f(\Gamma _0)=[-1,1]$, that is, $\Im(f(z))=0$ for all $z$ in the proper subarc $\Gamma _0$. Modify this function appropriately for your purpose.

The only remaining task is to check the continuity of $f$ on $\bar{D}$, which is obvious.