Holomorphic function multiplied by real number

Question

I would like to ask if a holomorphic function (say $z$) multiplied by some real constant is still a holomorphic function. It seems a bit obvious but I'm searching for a good argument.

Thanks

Answer 1

Alternatively: $f$ is holomorphic if and only if $f$ is complex-differentiable everywhere in its domain, if and only if the limit $$\lim_{z \to z_0} \frac{f(z) - f(z_0)}{z - z_0}$$ exists for all $z_0$ in the domain. Now, applying a simple property of limits, note that the limit $$\lim_{z \to z_0} \frac{k \cdot f(z) - k \cdot f(z_0)}{z - z_0} = k \cdot \left( \lim_{z \to z_0} \frac{f(z) - f(z_0)}{z - z_0} \right)$$ also always exists, for all $k \in \mathbb C$.

Answer 2

Yes. Recall the Cauchy-Riemann equation: A function $$f:\Bbb C\to\Bbb C,\quad f(x+iy)=u(x,y)+i\,v(x,y)$$ is holomorphic if and only if $u$ and $v$ are continuously differentiable and satisfy $$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}\quad\text{and}\quad\frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y} .\tag{1}$$ Now multiplying $f$ by a constant means multiplying both $u$ and $v$ by the same constant, and hence the result will still satisfy (1). (The constants will cancel on both sides of each equation.)