show $f$ complex analytic is constant

Question

$U\subseteq \mathbb{C}$ open and connected, $f : U \rightarrow \mathbb{C}$ complex analytic function.

Suppose $f(z) = u(x, y) + iv(x, y)$, $z = x + iy$, and $u$ and $v$ are real-valued functions.

Suppose that $v$ only takes integer values on $U$.

Using the Cauchy–Riemann equations and the fundamental theorem of the calculus, show $f$ is constant.

I'm not really sure what is meant by "$v$ only takes integer values on $U$". Does it mean that $v(x,y)=c$ for all $x,y \in U$ or does it mean that $v(x,y)=A(x)B(y)$ such that it equals an integer for all $x,y$?

I'm also just unsure in general how to use Cauchy Riemann and FTC to show f is constant.

Answer 1

You also need the assumption that $U$ is connected. First, note that $v$ must be continuous. Since it has integer values, it is constant. Then, using Cauchy-Riemann equations,

$$ \partial_x u = \partial_y v = 0$$

so

$$f'(x+iy) = \partial_x u (x,y) + \partial_x v (x,y) = 0$$

everywhere on $U$.

Note that connected implies path-connected. Fix $z_0 \in U$. For $z \in U$, choose a path $\sigma: [0,1] \to U$ from $z_0$ to $z$. Then,

$$f(z) - f(z_0) = \int_\sigma f'(z) dz = 0 $$

Answer 2

I'm going to assume $U$ is connected for this exercise.

Suppose that $f = u(x,y) + iv(x,y)$ only takes integer values, that means for every element $e$ of the set $U$ $f(e)$ is an integer.

We have that $f$ is complex analytic i.e. at all point $u_i =(x,y)$ in the set $U$ $f$ satisfies the cauchy riemann equations given as:

$$ \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \\ \frac{\partial u}{\partial y} = - \frac{\partial v}{\partial x}$$

So now here is our proof by contradiction. Suppose $f$ is NOT constant. That is we can find two points $u_1, u_2 \in U$ such that $f(u_1)\ne f(u_2)$ now because $U$ is connected there exists a path $P$ from $u_1$ to $u_2$ (obviously $u_1$ and $u_2$ are the two endpoints of this path). Let $u_{1,k}$ be a sequence of points in $P$ that starts with $u_{1,0} = u_1$ and has a limit point $L_1$ (still on the path) with the property that for any $\epsilon >0$ $f(L_1+\epsilon) = f(u_2)$ and $f(u_{1,k})= f(u_1)$. Similarly we can define $u_{2,k}$ starting from the other endpoint and going in the other direction.

It's clear that that at the point $L_1$ our function ceases to be continuous let alone complex differentiable. Or you could say, the Cauchy Riemann Equations are not well defined. Since it is an abrupt jump.

This contradicts one of our assumptions, and so we are forced to conclude that $f$ is not constant.