If $|f|+|g|$ is constant on $D,$ prove that holomorphic functions $f,~g$ are constant on $D$.

Question

Let $D\subseteq \mathbb{C}$ be open and connected and $f,~g:D \rightarrow \mathbb{C}$ holomorphic functions such that $|f|+|g|$ is constant on $D.$ Prove that $f,~g$ are constant on $D.$

Attempt. I noticed first that this problem is already set before, and is considered to be duplicate (If $|f|+|g|$ is constant then each of $f, g$ is constant) and we are sent directly to problem sum of holomorphic functions. However, the last reference deals with the sum $|f|^2+|g|^2$, which I do not see how is connected to the sum $|f|+|g|$, as stated in our title (for example, the equality $|f|^2+|g|^2=(|f|+|g|)^2-2|f||g|$ is not helpful here).

Thanks in advance!

Answer 1

By the maximum modulus principle $|f(z)|$ has no local minimum and maximum except at its zeros. So if $f(a)=0$ then $a$ is a local minimum of $|f(z)|$ so it is a local maximum of $|g(z)|$ and $g(a) = 0$, i.e. $f(z) = g(z)= 0$.

$\implies$ If $f(z)$ is non-constant then $f(z),g(z)$ have no zeros, and we can look at the holomorphic functions $F(z) = \log f(z), G(z) = \log g(z)$ and $H (z) = e^{Re(F(z))}+e^{Re(G(z))}=C$. Differentiating : $$\partial_x H(z) = Re(F'(z)) e^{Re(F(z))}+Re(G'(z))e^{Re(G(z))}=0 $$ $$ \partial_y H(z) = Im(F'(z)) e^{Re(F(z))}+Im(G'(z))e^{Re(G(z))}=0$$

so that $$0= F'(z)e^{Re(F)}+G'(z)e^{Re(G(z))}=(F'(z)-G'(z))e^{Re(F(z))}+G'(z)(e^{Re(F(z))}+e^{Re(G(z))})$$ $$ = (F'(z)-G'(z))e^{Re(F(z))}+C G'(z)$$ Finally $(G'(z)-F'(z))e^{Re(F(z))}=C G'(z)$ means that $(G'(z)-F'(z))e^{Re(F(z))}$ and $e^{Re(F(z))}$ is holomorphic, i.e. $Re(F(z))$ and $F(z),f(z)$ are constant.

Answer 2

Here is a proof that relies heavily on the open mapping theorem.

Since the zeros of $f,g,f',g'$ are isolated, we can find some $z_0$ such that $f,g,f',g'$ are non zero in a neighbourhood of $z_0$.

Suppose $z$ lies in this neighbourhood of $z_0$.

If we let $\phi(t) = {1 \over 2}|f(z+th)|$, it is straightforward to show that $\phi'(0) = { \operatorname{re}(\overline{f(z)} f'(z) h ) \over |f(z)|}$, hence since $|f(z)| + |g(z)|$ is constant we obtain \begin{eqnarray} { \operatorname{re}(\overline{f(z)} f'(z) h ) \over |f(z)|} + { \operatorname{re}(\overline{g(z)} g'(z) h ) \over |g(z)|} &=& \operatorname{re} \left[ \biggl( { \overline{f(z)} f'(z) \over |f(z)|} + { \overline{g(z)} g'(z) \over |g(z)|} \biggr) h\right] \\ &=& 0 \end{eqnarray} for all $h$.

Hence, we have ${ \overline{f(z)} f'(z) \over |f(z)|} + { \overline{g(z)} g'(z) \over |g(z)|} = 0$ and so ${f'(z) \over g'(z)} = -{|f(z)| \over \overline{f(z)} } {\overline{g(z)} \over |g(z)|}$, in particular, $| {f'(z) \over g'(z)}| = 1$. Hence ${f'(z) \over g'(z)} = c$ for some constant and so ${g(z) \over f(z)} = - \overline{c} | {g(z) \over f(z)} | $. Hence the values $z \mapsto {g(z) \over f(z)}$ takes lie on the line $\{t \overline{c} \}_{t \in \mathbb{R}}$ and hence we must have ${g(z) \over f(z)} = d$, another constant. Hence $(1+|d|)|f(z)|$ is a constant and so $f$ is a constant.

Answer 3

Here is an answer that will help you to solve the problem the way that you already tried to do it.

First let's enumerate our equation \begin{eqnarray} \big|f\big| + \big|g\big| = c. \quad(1) \end{eqnarray}

It's easy to see that f,g can't be zero on $D$, otherwise if for example there was $z_0\in D$ such as $f(z_0)= 0$ then you have that $$\big|g(z_0)\big| = c$$ but you also have that $$\big|g(z)\big|\leq c \quad \forall z\in D$$ so from maximum modulus principal you have that $g=c$ on $D$ so (1) gives us that $f = 0$ on $D$. So we have that there exists some holomorphic functions on $D$, $f_1,g_1$ such us \begin{eqnarray} &f_1^2=f \\ &g_1^2=g \end{eqnarray} (Theorem 13.11 page 274 from Real and Complex Analysis by Rudin). Now you have what you were looking for, you have that $$\big|f_1\big|^2+\big|g_1|^2 = c \quad (2)$$ and now all you need is to prove is that $f_1,g_1$ are constant. Now you can continue as that link you mention, but there is one other way to do it. We have that $f_1,g_1$ are holomorphic so they have a powerseries expansion \begin{eqnarray} &f_1(z) = \sum_{n=0}^{\infty}a_nz^n \\ &g_1(z) = \sum_{n=0}^{\infty}b_nz^n \end{eqnarray} so if we integrate on some circle $\gamma$ with centre $0$ and radious $r$ (I suppose that $\;D = D(0,R)\;$ and $\;0<r<R\;$) equation (2), Parseval's formula gives us that \begin{eqnarray} &\frac{1}{2\pi }\int_{\gamma}|f_1|^2dz + \frac{1}{2\pi }\int_{\gamma}|g_1|^2 dz =\frac{1}{2\pi}\int_{\gamma}cdz \Longrightarrow \\ &c_1 = \sum_{n=0}^{\infty}|a_n|^2r^{2n}+\sum_{n=0}^{\infty}|b_n|^2r^{2n} = \sum_{n=0}^{\infty}\Bigl(|a_n|^2+|b_n|^2\Bigr)r^{2n} \quad \forall r\in(0,R) \end{eqnarray} where $\; c_1\in \mathbb C\;$ is a constant, so all $a_n,b_n = 0$ for $n\geq 1$, and you have the result.