# Multiplication of analytic functions

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.

• Holomorphic functions are analytic. Jan 26, 2014 at 21:06
• What is your definition of analytic?
– lhf
Jan 26, 2014 at 21:10
• Erm.. Function is differetiable in some open disc D Jan 26, 2014 at 21:15
• Since your definition is that a function is analytic if it's differentiable in an open disc, note that the product of differentiable functions is differentiable.
– user61527
Jan 26, 2014 at 21:16
• Note however that analyticity usually refers to the function being smooth and that it is locally represented by a power series. For complex functions this is equivalent to being locally (complex) differentiable, however that is a major theorem.
– J.R.
Jan 26, 2014 at 21:51

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.

• Thanks for comment lhf, this is what was bugging me, and you guessed perfectly. Thanks for other efforts of course as well. But.. problem was with different definitions in different lang.. Analyyttinen(fi) isnt analytic, it is holomorphic! Retrospectively TooOld also guessed right. Jan 27, 2014 at 18:04
• Please use Latex for math. Many authors use analytic and holomorphic as synonyms, by the way. Jan 27, 2014 at 18:07
• Promise, next time - Latex, sorry for this mess, future reader. Jan 27, 2014 at 18:16
• You know you can edit posts, right? Jan 27, 2014 at 18:16
• actually you are right, but not today - tomorrow ish(latest being thursday) i will reformat this mess. Newbie,exited and in zeitnot, very naughty combination. Jan 27, 2014 at 18:21

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 ${\mathbb R}$.