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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 $\Omega$?

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    $\begingroup$ 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. $\endgroup$ Commented Sep 21, 2020 at 12:03

2 Answers 2

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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.

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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 $g$?

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  • $\begingroup$ 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 $0$)? $\endgroup$
    – user824237
    Commented Sep 21, 2020 at 12:47
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    $\begingroup$ 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 $\endgroup$ Commented Sep 21, 2020 at 12:59

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