Holomorphic function and an open disc

[Stephen88]

Now the function f is holomorphic in an open disc U and that Re( f ) is
constant in U. I'm trying to show that
1)f must be constant in U.
2) the essential property of the disc U that it used here
3) an example of an open set U for which the conclusion fails.

Let f=u+vi where u is a constant.Since f is holomorphic by the Cauchy–Riemann equations->
u_x=v_y and u_y=-v_x but since u is a constant u_x=u_y=0 => 0=v_y =-v_x...therefore f is constant.
The disc U has to be open,as in:U(a,r)={z:|z-a|<r}.
Is this correct?What should I do for the last part?
Thank you

 

[PaulRS]

Note that your proof of 1) actually uses the fact that the disc is connected as well as the fact that it is open.

$U$ open implies that given $p\in U$ we must have $\varepsilon > 0$ such that $B_{\varepsilon} (p) \subseteq U$. But $B_{\varepsilon} (p)$ is convex , thus $v(x) - v(p) = \bigtriangledown v (c) \cdot (x-p) = 0$ for some $c\in B_{\varepsilon} (p)$ (*) and so $v(x) = v(p)$ for all $x\in B_{\varepsilon} (p)$. Thus $u$ is locally constant, and each set $v^{-1}\left(\{a\}\right)$ is open for $a\in \mathbb{R}$, so if $U$ were connected, $v$ can take only one value (because otherwise our connected set would be the union of 2 or more disjoint non-empty open sets, which is a contradiction).

So if you want to solve (3) look at a disconnected set (for instance, the union of 2 disjoint open balls).

Here: $B_{\varepsilon}(p) := \{z \in \mathbb{C} : |z-p| < \varepsilon\}$

(*) Mean value theorem for the function $g(t) = v\left(p\cdot (1-t) + x\cdot t\right) $ , $g: [0,1] \to \mathbb{R}$. This makes sense since we are working on the convex set $B_{\varepsilon} (p)$.

 

[ThePerfectHacker]

I want to add something to what PaulRS said.

Paul showed that f must be locally-constant. Here is a generalization of his statement.

Here is a purely topological exercise. Show that if f is locally constant on U then it must mean that f is constant on the connected component of U. In particular if U is connected then f is constant on U.

 

[AlexYoucis]

I want to add something to what PaulRS said.

Paul showed that f must be locally-constant. Here is a generalization of his statement.

Here is a purely topological exercise. Show that if f is locally constant on U then it must mean that f is constant on the connected component of U. In particular if U is connected then f is constant on U.

This is a good exercise because of how often it shows up in things like complex analysis/differential geometry. A similar exercise which is much simpler, but actually comes up even more than the exercise TPH suggested is the trivial matter that continuous maps from connected spaces to discrete spaces are constant. Useful for proving that different branches of the logarithm differ from each other by a constant multiple of $2\pi i$.

 

[blue_raver22]

for your response.

1) Yes, your explanation for why f must be constant in U is correct. Since the real part of f is constant in U, it follows that the imaginary part of f must also be constant, and therefore f itself is constant in U.

2) The essential property of the disc U that is used here is its openness. This means that every point in U has a neighborhood contained within U. This is important because it allows us to use the Cauchy-Riemann equations, which require the function to be differentiable at every point in U.

3) An example of an open set U for which the conclusion fails is the unit circle, defined as U={z:|z|=1}. In this case, the function f(z)=iz is holomorphic in U and has a constant real part (Re(f) = 0), but it is not constant in U. This is because the unit circle is not an open set, as it does not contain any points in its interior. Therefore, the Cauchy-Riemann equations cannot be applied.