Prove that if $f$ and $g$ are analytic at $w$, then so is $fg$

Question

Prove that if $f$ and $g$ are analytic at $w$, then so is $fg.$

My main attempt was using the Cauchy-Riemann equations on the product in this manner but this did not work out. My thinking:

\begin{align} f =u_1+iv_1, \\ g =u_2+iv_2, \end{align}

then $fg=(u_1u_2-v_1v_2)+i(u_1v_2+v_1u_2)$. Set $\phi= u_1u_2-v_1v_2, \ \pi = u_1v_2+v_1u_2$. I need to prove $$ \frac{\partial \phi}{\partial x} = \frac{\partial \pi}{\partial y} \\ \frac{\partial \phi}{\partial y} = -\frac{\partial \pi}{\partial x} $$

I tried other things as well but nothing is working for me.

Answer 1

Here is an easy way to see it, using one of the definitions. A function is analytic if its derivative with respect $\overline w$ is zero. We have $$ \frac{\partial fg}{\partial \overline w}(w_0)=f(w_0)\frac{\partial g}{\partial \overline w}(w_0)+g(w_0)\frac{\partial f}{\partial \overline w}(w_0)=0, $$ because $f$ and $g$ are analytic at $w=w_0$.

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Following your approach: \begin{align} \frac{\partial \phi}{\partial x}& =\frac{\partial (u_1u_2-v_1v_2)}{\partial x}= \\ & =\frac{\partial u_1}{\partial x}u_2 + u_1 \frac{\partial u_2}{\partial x} - \frac{\partial v_1}{\partial x}v_2 - v_1 \frac{\partial v_2}{\partial x} \\ & = \frac{\partial v_1}{\partial y}u_2 + u_1 \frac{\partial v_2}{\partial y} + \frac{\partial u_1}{\partial y}v_2 + v_1 \frac{\partial u_2}{\partial y}= \\ & = \left( \frac{\partial v_1}{\partial y}u_2 + v_1 \frac{\partial u_2}{\partial y}\right) + \left( u_1 \frac{\partial v_2}{\partial y} + \frac{\partial u_1}{\partial y}v_2\right) = \\ & = \frac{\partial (u_2v_1 + u_1v_2)}{\partial y} = \\ & = \frac{\partial \pi}{\partial y}. \end{align} and similarly for $\dfrac{\partial \phi}{\partial y} = -\dfrac{\partial \pi}{\partial x}$.

Answer 2

Whatever proof you know for $(fg)' = f'g+fg'$ from elementary calculus will work nearly verbatim in this situation.

Answer 3

The straightforward argument works too. That the power series for $f$ and $g$ at a point $a$ both converges absolutely within at least the smallest of two radii of convergence, so their Cauchy product converges to $fg$.