# The real part of a holomorphic function and its link to the function being constant

(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.

• What is that number $a$ in the solution of (a)? – José Carlos Santos Jan 27 at 18:38
• 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 $f$. – Zachary Jan 27 at 18:42
• Yes, $f$ is purely imaginary on $S^1$. But why is it constant there? – José Carlos Santos Jan 27 at 18:47
• 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. – Zachary Jan 27 at 18:51
• The thing to build on is the fact that real and imaginary parts of a holomorphic function are harmonic. – Daniel Fischer Jan 27 at 19:17

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