Composition of analytic function with non analytic function.

Question

I have the following question.

If I have two complex valued functions, $f,g$. I know that if both are analytic then $f\circ g$ is analytic (if domain/codomain matches). Now I thought about if $g$ is not analytic, can we then immediately deduce that $f\circ g$ is not analytic neither? If not is there a counterexample?

Because if I consider for example $\sin(\bar z)$ then $\bar z$ is not analytic and also $\sin(\bar z)$ isn't but $\sin$ is analytic.

Thanks for your help.

Answer 1

Take $f$ to be the constant funciton equal to $0$ and $g$ to be your favourite non-analytic function. Then, $f \circ g$ is constant equal to $0$, and hence analytic.

Answer 2

For each $z\in\Bbb C$, let $g(z)$ be a square root of $z$. Then $g$ is not analytic. Acutally, it's not even continuous. But if $f(z)=z^2$, then $(f\circ g)(z)=z$, for each $z\in\Bbb C$, and therefore $f\circ g$ is analytic.