Product rule for complex differentiation

Statement

Complex-differentiable at a point

Suppose $U\subset \mathbb{C}$ is an open subset and $f,g:U\to \mathbb{C}$ are functions. Suppose $z_0\in U$ is a point such that $f,g$ are both complex-differentiable at $z_0$. Define $fg:U\to \mathbb{C}$ as:

$fg:=z\mapsto f(z)g(z)$

Then $fg$ is complex-differentiable at $z_0$ and:

$(fg{)}^{\prime }(z_0)=f^{\prime }(z_0)g(z_0)+f(z_0)g^{\prime }(z_0)$

For holomorphic functions

Suppose $U\subset \mathbb{C}$ is an open subset and $f,g:U\to \mathbb{C}$ are holomorphic functions. Then the function $fg:U\to \mathbb{C}$ given by:

$fg:=z\mapsto f(z)g(z)$

is also a holomorphic function, and for any $z\in U$, we have:

$(fg{)}^{\prime }(z)=f^{\prime }(z)g(z)+f(z)g^{\prime }(z)$