# A Complex Function is Constant if It Satisfies One of These Properties

I am given that a complex function $$f(z) = u(x,y) + iv(x,y)$$ is analytic in a region $$\Omega$$. I am also given a list of conditions and must show that if $$f$$ satisfies any one of those conditions in $$\Omega$$, then $$f$$ is constant in $$\Omega$$. These are the conditions that I am having trouble with:

a. $$f(\Omega) = \{f(z) : z \in \Omega\}$$ is a subset of a circle.

b. $$u^n(x,y) = v(x,y)$$ for some $$n \in \mathbb{N}$$.

c. $$Re(f)$$ is analytic on $$\Omega$$.

My thoughts for part a: is this property saying that f is bounded by the subset of the circle? If so, can I apply Liouville's Theorem and say that since f is bounded, it must be constant?

For part b., my approach was to substitute $$u^n(x,y)$$ in for $$v(x,y)$$ in the function, and then use the Cauchy Riemann equations (since $$f(z)$$ is analytic). Calculating the partial derivatives gave me $$u_x = nu^{n-1}u_y$$ and $$u_y = -nu^{n-1}u_x$$. I am not sure if I did the partial differentiation correctly, though, and I'm not sure where to go from here.

For part c., I understand what it means for a complex function to be analytic, but what does it mean for its real part to be analytic? And how does that show that $$f$$ is constant on the region $$\Omega$$?

• You need to put \ in front of brackets to have them appear in $\rm\LaTeX$. The reason being is that brackets are used for macros, so it will simply skip over them if there isn't a function to go with them for self-consistency. Commented Sep 21, 2020 at 12:03

A nice way to prove these results relies solely on the open mapping theorem. One can prove the following, stronger result:

Let $$G$$ be a $$\mathcal{C}^{1}(\mathbb{C},\mathbb{R})$$ function such that $$0$$ is a regular value of $$G$$ (in particular $$G^{-1}(0)$$ cannot contain an open set). If $$G(f)=0$$ on a region $$\Omega$$, $$f$$ is constant.

The result is easy to prove: let $$f$$ be non constant. Then it is an open mapping, which contradicts the regularity of $$0$$ for $$G$$ .

This result suffices to prove a&b. For $$c$$, if $$\text{Re}(f)$$ is analytic it must be constant by the open mapping theorem, and this implies $$\text{Re}(f(\Omega))=\{c\}$$ and again by the open mapping theorem we obtain the result.

Your argument for (a) would be fine if $$f$$ was defined on all of $$\mathbb{C}$$. $$\Omega$$ could be the open unit disk, in which case your argument fails. Hint: no subset of the unit circle can be open in $$\mathbb{C}$$.

For (b), you're on the right track. You have $$u_x = nu^{n-1} u_y$$ and $$u_y = -nu^{n-1}u_x$$ which after substitution gives $$u_x = -n^2 u^{2n-2} u_x$$ or $$u_x(1+n^2 u^{2n-2}) = 0$$ (and similarly for $$y$$). Note that $$1+n^2 u^{2n-2}$$ is positive..

For (c), if $$\operatorname{Re}(f)$$ were (presumably) complex analytic, then it would satisfy the Cauchy-Riemann equations. Writing $$g(x,y) = \operatorname{Re}(f(x,y))$$, what would the Cauchy-Riemann equations lead you to concluding about $$g$$?

• I'm still confused by part a. I see that the subset will be closed and so $f$ is bounded by that closed subset, but I'm not seeing how that shows that we will have $f$ be constant. Does the closed subset imply something about the values of the partial derivatives (i.e. that they are $0$)?
– user824237
Commented Sep 21, 2020 at 12:47
• I can't give you a better hint than what is there already. I almost spell the result you need to reference out for you :P Commented Sep 21, 2020 at 12:59