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Suppose $h(z)$ is a complex function. I have noticed that $h(z) = f(z)\cdot g(z)$, where $f(z)$ and $g(z)$ are non-holomorphic. Can $h(z)$ be holomorphic?

Can a similar statement be made, if $f(z)$ is holomorphic, but $g(z)$ is not?

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Yes, consider $\bar z \cdot \dfrac 1{\bar z} = 1$. However, if $f$ is holomorphic and $h$ is holomorphic, then $g=h/f$ is holomorphic wherever $f\ne 0$.

Most of the time, you can make a similar statement when $f$ is holomorphic while $g$ is not, but you have to modify it a bit to rule out silly cases like $f(z)=z^2$ and $g(z)=\frac1z$ when $z\neq0$ and $g(z)=0$. In this case, $h(z)=z$ is holomorphic, but in reasonable cases, $h$ will not be holomorphic.

Specifically, When $f$ is holomorphic and $g$ is not *meromorphic*, you can conclude have $h$ is not holomorphic.

Assume $h$ is holomorphic. Since $1/f$ is holomorphic except at the zeroes of $f$, which must be isolated, we have that $h\cdot(1/f)=g$ is holomorphic except for some poles or removable singularities (depending on whether the zeroes of $h$ cancel those of $f$). This shows $g$ is meromorphic.

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