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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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    how do i use identity theorem to show that? well, i know if f and f¯ are holomorphic then f is constant, then im done with the first part. but what about the second part. g=0?2012-05-19

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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.

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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 $G$.

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    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.2016-05-21