If $f(z)$ and $g(z)$ are not analytic, then $fg$ can be analytic?

Question

If $f(z)$ and $g(z)$ are not analytic, then $f(z)g(z)$ can be analytic?

I guess this is false. Let’s take $f(z)=\frac{z}{z+1}$ and $f(z)=\frac{z+1}{z}$. Then obviously $f(z)g(z)=1$ but this function is not defined at $z=\{ 0, -1 \}$.

But how can I prove it in general?

Answer 1

$|z|$ and $\frac {z^{2}} {|z|}$ are not analytic but their product is. [ The second one is defined to be $0$ at $0$. That makes both functions continuous].

Answer 2

Be $$f(x)=\begin{cases} 1 & x\in\mathbb Q\\ -1 & x\notin\mathbb Q \end{cases},\quad g(x)=\begin{cases} -1 & x\in\mathbb Q\\ 1 & x\notin\mathbb Q \end{cases}$$ Clearly, neither of those functions is analytic anywhere in $\mathbb R$. However $f(x)g(x)=-1$ for all $x$, and the constant function is perfectly analytic.

Answer 3

Take $f(z)=e^{|z|}$ and $g(z)=e^{-|z|}$. Then $f(z)g(z)=1$.