# Show either $f$ is constant or $g(z)=0$ for all $z$ in the region

let $G$ be a region, and $f$ and $g$ be holomorphic function on $G$. if $\bar{f}\cdot g$ is holomorphic, show that either $f$ is a constant or $g(z)=0$ for all $z$ in $G$.

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What does f(bar)*g mean? – Andrés E. Caicedo May 19 '12 at 23:09
Have you tried using the Cauchy-Riemann equations? The result should fall out of writing f and g in terms of their real and imaginary parts and fiddling around a bit. @Andres: bar denotes the complex conjugate. You can't tell if he means the conjugate of f or evaluating f on the conjugate of z. – Connor May 19 '12 at 23:13
"You can't tell if he means the conjugate of f or evaluating f on the conjugate of z." And therefore the question. – Andrés E. Caicedo May 19 '12 at 23:16
And what does this have to do with analytic geometry? – Andrés E. Caicedo May 19 '12 at 23:18
thx for the reply! f(bar) is f conjugate, and it says f(bar) time g is holomorphic. and the hint is that we suppose to use the identity theorem. – complexmath May 19 '12 at 23:19

If $g(z)$ is not identically zero, you can find an open set $U$ on which $g(z)$ is nonzero. Thus on $U$, $\bar{f} = \bar{f} g \times {1 \over g}$ is analytic. So ${1 \over 2}(f + \bar{f})$ = $Re(f)$ and $Im(f) = {1 \over 2i}(f - \bar{f})$ are analytic functions on $U$. Applying the Cauchy-Riemann equations to these two functions gives that $Re(f)$ and $Im(f)$ are constant on $U$, so the same is true for $f$. Since $f$ is equal to a constant on an open subset of your region, it's equal to that constant everywhere.
I think we have to assume $G$ is connected otherwise this does not work : let $G = G_1 \cup G_2$ with $G_1$ and $G_2$ disjoint open sets, set $f = 1_{G_1}$ (ie $f(z)$ is $1$ if $z \in G_1$ and $0$ otherwise), $g = 1_{G_2}$. Then $f$, $g$ and $\overline{f}g$ are holomorphic but $f$ is not constant, and $g$ is not zero.
We now assume $G$ is connected. Denote $h = \overline{f} g$. Let $\Omega \subset G$ be the open subset of points on which $g$ does not vanish. Assume $\Omega$ is non empty (otherwise $g = 0$ and we're done). Then $f$ and $\overline{f} = \frac{h}{g}$ are holomorphic on $\Omega$, so $f$ is constant on $\Omega$ (using the Cauchy-Riemann equation). And because $G$ is connected, the identity theorem shows that $f$ is constant on $G$.
I know this is an old question but I have some doubts with your answer: You consider $\Omega$ the open subset of points where $g$ does not vanish. $f$ and $\overline{f}$ are holomorphic on $\Omega$, so you've affirmed that by C-R one deduces that $f$ is constant there. By Cauchy-Riemann you get that if $f=u+iv$, then $\nabla u=0=\nabla v$, if $\Omega$ was connected then you can affirm that $u$ and $v$ are constant, but this is not necessarily the case. – user16924 May 21 at 17:39
So. instead of $\Omega$, maybe one could pick a point $z_0$ in which $g$ does not vanish and then apply all these arguments of the Cauchy-Riemann equations and etc. in a ball $B(z_0,\epsilon)$ in which $g$ does not vanish, and since a ball is a connected set, then there you can assure that $f$ has to be constant. – user16924 May 21 at 17:39