Let $U=\{z\in\mathbb{C}\mid 1\leq \lvert z\rvert \leq 2\}.$ Prove that there is no non-constant holomorphic function on $U$ with $\operatorname{Re} f(z)=C_1$ for $\lvert z\rvert=1$ and $\operatorname{Re}f(z)=C_2$ for $\lvert z\rvert=2.$
Using maximum modulus principle for $e^{f(z)}$ and $e^{-f(z)}$ one can show that $\min\{C_1,C_2\}\leq\operatorname{Re}f(z)\leq \max\{C_1,C_2\},$ but since $f$ can fail to be entire this doesn't give a contradiction.
Thank you.
Best Answer
Consider the harmonic function $u=\Re f-C_1-\frac{C_2-C_1}{\log 2}\log |z|$.
Then $u=0$ on the boundary of $U$ hence $u=0$ on $U$ or $\Re f=C_1+\frac{C_2-C_1}{\log 2}\log |z|$
Assuming $C_1 \ne C_2$ we get that if $g=e^f, |g|=A|z|^B, |g|^{\frac{1}{B}}=A_1|z|, A, A_1>0, B \ne 0$
Hence $|\frac{1}{z}e^{\frac{f}{B}}|=A_1$ so by maximum modulus $e^{\frac{f}{B}}=\alpha A_1z, |\alpha|=1$ and diferentiating we get $z{\frac{f'}{B}}=1$ on $U$ or ${\frac{f'}{B}}=\frac{1}{z}$.
But now integrating on any circle inside $U$ we get $0=2\pi i$ and that is a contradiction!
Hence $C_1=C_2, B=0, f$ constant so done!