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:

$(fg)'(z_{0})=f'(z_{0})g(z_{0})+f(z_{0})g'(z_{0})$

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:

$(fg)'(z)=f'(z)g(z)+f(z)g'(z)$