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(a) Consider $f$ holomorphic on $\overline{B}(0,1)$ and $\Re(f)=0$ for $|z|=1$. Show that $f$ is constant.

(b) Let $M\in\mathbb{R}$. Suppose that $\Re(f)=M$ for $|z|=1$. Is $f$ still constant? Prove or give a counterexample.

(c) Suppose now that $\Re(f)\le 0$ for $|z|=1$. Is $f$ still constant? Prove or give a counterexample.

My attempt:

(a) We know that $\Re(f)=0$ on the boundary of the compact disk: $f(z)=ia, a\in\mathbb{R}$ for $|z|=1$. By the maximum modulus principle, we can conclude that $\max_{z\in\overline{B}(0,1)}|f(z)|=\max_{|z|=1}|a|=|a|$. So, $\forall z\in\overline{B}(0,1): |f(z)|\le |a|\Rightarrow \sqrt{ (\Re(f))^2+a^2}\le |a|$. The LHS is at least equal to $|a|$, implying that $|f(z)|=|a|$ for all $z\in\overline{B}(0,1)$. This however implies that $f$ is constant, because $|f|$ reaches its maximum on $B(0,1)$ (it is constant).

(b) The given information basically means that the unit circle is mapped to a vertical line. The question becomes "is the imaginary part of $f$ also a constant?". If so, then $f$ is constant. I tried to apply the maximum modulus principle but the upper bound becomes very messy ($\max_{z\in \overline{B}(0,1)}|f(z)|\le M+\max_{|z|=1}|\Im(f)|$). So, I tried to reduce the problem to case (a): consider $f(z)-M$. This function is holomorphic on $\overline{B}(-M,1)$ and $\Re(f-M)=0$ for $|z-M|=1$. This shows that $f-M$ is constant (via (a)) and therefere $f$ is still constant. Is this OK?

(c) Just by observing this question and the previous ones, I can see that this should be false (or true in cases (b) was false; otherwise (b) and (c) would be considered as one problem, right?). But since I didn't put any constraints on $M$ in (b), I feel like my previous reasoning was incorrect (because $M\le 0$ is also possible).

Could someone please point out any mistakes in (a) and (b), and give a starting point for (c)? Thanks.

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  • $\begingroup$ What is that number $a$ in the solution of (a)? $\endgroup$ – José Carlos Santos Jan 27 at 18:38
  • $\begingroup$ Since $\Re(f)=0$ on the boundary, the function will be constant for $|z|=1$. But constant and pure imaginary since the real part is zero. The real number $a$ is the imaginary part of $f$. $\endgroup$ – Zachary Jan 27 at 18:42
  • $\begingroup$ Yes, $f$ is purely imaginary on $S^1$. But why is it constant there? $\endgroup$ – José Carlos Santos Jan 27 at 18:47
  • $\begingroup$ I think I've applied a theorem, when its condition weren't fulfilled; i.e. if $f=u+iv$ is holomorphic on an open domain $\Omega$ and $u=0$ or $v=0$, then $f$ is constant. But the boundary is not an open domain, meaning that (a) and (b) are guaranteed to be incorrect. $\endgroup$ – Zachary Jan 27 at 18:51
  • $\begingroup$ The thing to build on is the fact that real and imaginary parts of a holomorphic function are harmonic. $\endgroup$ – Daniel Fischer Jan 27 at 19:17
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There is a nice exercise that will give you a): Suppose $g$ is holomorphic on $\overline B (0,1),$ with $|g|=1$ on the unit circle. Then either $g$ is constant or $g$ has a zero in $B(0,1).$ Basic idea of proof: If $g$ has no zero in $B(0,1),$ then $1/g$ satisfies the same hypotheses. Apply the the maximum modulus theorem to both $g, 1/g$ to see $g$ must be constant.

To get a), consider $g=e^f.$

b) I'm not sure what you're doing here. Doesn't $f-M$ satisfy the hypotheses in a)?

c) Consider $f(z)=z-2.$

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