$\lim_{z\to\\z_0} f(z)g(z)$=0.
This is for complex analysis, and I know how to do this argument for regular epsilon calculus arguments. We want to say since $\lim_{z\to\\z_0} f(z)$=0 then we can take
$|f(z)-0|<\frac{e}{M}$ whenever $0<|z-z_0|\delta$. Okay I know we need to have an argument where we take $\delta=\min\{blah,blah\}$ maybe? But the underlying algebra is,
$|f(z)|*|g(z)| = |f(z)*g(z)-0|<\frac{e}{M}$*M =$\epsilon$
Best Answer
Let $\epsilon > 0$ be given. Since $f(z) \to 0$, there exists a $\delta > 0$ such that
$$|z - z_0| < \delta \implies |f(z)| < \frac{\epsilon}{M}$$
Now what can you say about $|f(z)g(z)|$ for $|z - z_0| < \delta$?