Let $f : \mathbb{C}\setminus$ {$0$} $\to \mathbb{C}$ be an analytic function with a simple pole of order $1$ at $0$ with residue $a_1$. Let $g : \mathbb{C} \to \mathbb{C}$ be analytic with $g(0)\neq 0$.calculate for $r>0$
$$\frac{1}{2\pi i} \int_{|z|=r}{f(z)g(z)dz}$$
my thoughts:
the answer will be $Res(f(z)).g(0)$ that is $a_1g(0)$. am I right?
Quick check:
We can write
$$f(z)=\frac{h(z)}{z}\;\;,\;\;\text{with analytic}\;\,h(z)$$
and so
$$\frac{1}{2\pi i}\int\limits_{|z|=r}f(z)g(z)\,dz=\frac{1}{2\pi i}\int\limits_{|z|=r}\frac{h(z)g(z)}{z}dz\stackrel{\text{Cauchy's Theorem}}=h(0)\cdot g(0)$$
But, of course, $\,h(0)=a_1\ldots$
Hint: Since $f$ has a simple pole at $z=0$ with residue $a_1$, then you can write
$$ f(z)=\frac{a_1}{z}+h(z), $$
where $h(z)$ is an analytic function.
