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I would like to know if my reasoning is correct. I tried to prove the following :

Let $\Omega \subseteq \Bbb C$ a connected open set, $f : \Omega \to \mathbb C$ a holomorphic function, such that $u = \Re(f)$ is constant on some ball $B \subseteq \Omega$, then $f$ is constant on $\Omega$.

It is sufficient, by the identity theorem, to show that $f$ is constant on $B$.

If $f$ was not constant on $B$, then by using the open mapping theorem, $f(B) = \{c+i\Im(f)(z) \mid z \in B \}$ is open in $\Bbb C \cong \mathbb R^2$, where $c=\Re(f)(z) \in \Bbb R$ is a constant (on $B$). Then the projection $\text{pr}_1(f(B)) = \{c\}$ should be open in $\Bbb R$ which is not the case.

Therefore $f$ is constant on $B$ and on $\Omega$.

Any comment is welcome !

NB : if $u$ is bounded on $\Bbb C$, then my claim has already been answered here.

Community
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Alphonse
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2 Answers2

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I suggest that instead you use the Cauchy-Riemann equations to conclude that the imaginary part of $f$ has zero partial derivatives on $B$. Hence, $f$ is constant on $B$.

John B
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If $f(x,y)=u(x,y)+iv(x,y)$ is holomorphic and constant on $B\subset \Omega $, then $u_x=v_y=0$ and $u_y=-v_x=0$ (by Cauchy-Riemann) and thus $v$ is also constant on $B$. Therefore $f$ is constant on $B$.

Surb
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