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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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    I meant rather that $|f(z)|\equiv 1$. Did I make an error?2012-08-29

3 Answers 3

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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.2012-08-29
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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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    This is certainly a good idea if we assume $V$ is open, though.2012-08-29
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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.2012-08-29