Multiplication of analytic functions

Question

Suppose we have $f$ and $g$ and that both are analytic in neighborhood $D$.

Is $fg$ also analytic in $D$? (Afaik yes)

This is homework question, and to be honest - I am getting quite a mess with my approach by representing $f$ and $g$ as infinite series and getting Cauchy product.

Seems like there should be other approach, thanks a lot in advance.

Answer 1

I think i got it. Say $f(x+iy)=u_f(x,y)+iv_f(x,y)$ and $g(x+iy)=u_g(x,y)+iv_g(x,y)$

those should be in Cauchy Riemann relation, $du/dx=dv/dy$ respectively for f and g.. and I am too lazy to write second part of CR :-)

so $fg=(u_fu_g-v_fv_g)+i(u_fv_g+u_gv_f)$

where new(fg) $u'=(u_fu_g-v_fv_g)$ and $v'=(u_fv_g+u_gv_f)$

Now if CR works for new u and v, then it is holomorphic. new $\frac{\partial u'}{\partial x}=\frac{\partial u_f}{\partial x}u_g+u_f\frac{\partial u_g}{\partial x}-\frac{\partial v_f}{\partial x}v_g-\frac{\partial v_g}{\partial x}v_f$

new $\frac{\partial v'}{\partial y}=\frac{\partial u_f}{\partial y}v_g+\frac{\partial v_g}{\partial y}u_f+\frac{\partial u_g}{\partial y}v_f+\frac{\partial v_f}{\partial y}u_g=$(substituting related $\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$)=$-\frac{\partial v_f}{\partial x}v_g+\frac{\partial u_g}{\partial x}u_f-\frac{\partial v_g}{\partial x}v_f+\frac{\partial u_f}{\partial x}u_g$=new $\frac{\partial u'}{\partial x}$ from previous line. This part of CR works.. again i am too lazy and exited to write whole second part, but it should work as well, and product of to holomorphs is holomorph.

Answer 2

The usual proof of the product rule for derivatives only uses the continuity of the field operations. These are present in ${\mathbb C}$ as well as in ${\mathbb R}$.