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Let $V$ be a connected set in $\mathbb{C}$, and $f$ holomorphic on $V$ such that $$f(z)^2=\overline{f(z)}$$ on $V$.

I want to show that $f$ must be constant on $V$.

My attempt: As $$|f(z)|^2=f(z)\overline{f(z)}=|f(z)||\overline{f(z)}|=|\overline{f(z)}|,$$ it follows that the $\overline{f(z)}\equiv 1$ on $V$. Then it would follow that the imaginary part of $f$ must also be constant since it must be harmonically conjugate to $1$.

Is this ok?

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My \overline's are acting funny. –  Thelonius Aug 29 '12 at 1:14
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Here's a hint: can you prove that a function which is both holomorphic and anti-holomorphic (meaning that the conjugate is holomorphic) is constant? –  Akhil Mathew Aug 29 '12 at 1:17
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Your proof is essentially right. You should note the possibility that $f\equiv 0$ in $V$ as well. Also, alternatively, you could argue that $|f|\equiv 1$ implies $f$ is constant by the open mapping theorem. –  froggie Aug 29 '12 at 1:18
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I don't see how you got $\overline{f(z)}=1$. What I would do first is ask myself: which complex numbers have the property that the square of the number equals its conjugate? –  Gerry Myerson Aug 29 '12 at 1:27
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I meant rather that $|f(z)|\equiv 1$. Did I make an error? –  Thelonius Aug 29 '12 at 1:39

3 Answers 3

up vote 5 down vote accepted

Why isn't anyone taking my hint from the comments?

  1. If $z^2=\overline z$ then $z=0,1,e^{2\pi i/3}{\rm\ or\ }e^{4\pi i/3}$.

  2. If a continuous function takes on only finitely many values on a connected set, it's constant.

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Thanks, this is a good answer. I would have been happier if someone pointed out the flaw in my own argument (modulo my typo noted in the comments above), but this is certainly a nicer argument in any case. –  Thelonius Aug 29 '12 at 13:14

Since $f(z)$ is holomorphic, if $f=u+iv$, its real and imaginary parts satisfy Cauchy-Riemann equation, i.e. $$\tag{1}u_x=v_y\mbox{ and }u_y=-v_x.$$ Since $f(z)$ is holomorphic, $\overline{f(z)}=f(z)^2$ is also holomorphic. Since $\overline{f}=u-iv$, Cauchy-Riemann equation implies that $$\tag{2}u_x=-v_y\mbox{ and }u_y=v_x.$$ Combining $(1)$ and $(2)$, we have $v_y=0$ and $v_x=0$ on the connected set $V$, which implies that $v$ is constant on $V$. Since $v_y=0$ and $v_x=0$, we have $u_y=0$ and $u_x=0$ on $V$, which implies that $u$ is constant on $V$. Therefore, $f=u+iv$ is constant.

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It is not immediate to me that you can claim that $\overline{f(z)}$ is holomorphic on $V$ unless you know something about $V$ (such as being open, for example). For example, $f(z) = |z|$ equals the holomorphic function $z \mapsto 1$ on the unit circle, but is clearly not holomorphic. –  copper.hat Aug 29 '12 at 4:33
    
Since $f(z)$ is holomorphic on $V$ by assumption, $f(z)^2$ is also holomorhpic on $V$. By assumption, $\overline{f(z)}=f(z)^2$. So $\overline{f(z)}$ is holomorphic on $V$. Am I correct? –  Paul Aug 29 '12 at 4:53
    
Well, I don't think it follows just from equality on $V$, unless $V$ is, for example, open. See my example above of a holomorphic function equaling a non-holomorphic function on a particular set. –  copper.hat Aug 29 '12 at 4:55
    
This is certainly a good idea if we assume $V$ is open, though. –  Potato Aug 29 '12 at 6:24

Here is a proof that only requires $f$ to be continuous.

Since $|f(z)^2 |= |f(z) |^2=|f(z)|$ it follows that $|f(z)| \in \{0,1\}$ for all $z \in V$. Since $V$ is connected and $f$ is continuous, it follows that either $|f(z)| = 0$ on $V$ or $|f(z)| = 1$ on $V$.

If $|f(z)| = 0$ on $V$, then clearly $f$ is constant on $V$, so assume that $|f(z)| = 1$ on $V$.

Consider $f(z)^3 = |f(z)|^2$. Since $|f(z)| = 1$, we have $f(z)^3 = 1$ for $z \in V$. Since the cube roots of unity form a disconnected set, it follows that $f(z)$ is constant (equal to one of the cube roots of unity, of course) for $z \in V$.

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It occurs to me that this was essentially a long-winded way of following @Gerry Myerson's suggestion above. –  copper.hat Aug 29 '12 at 4:29

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