Finding formula for $Res(f_1f_2;z_0)$ Mmm so I'm given the problem;
If $f_1, f_2$ are functions with simple poles (degree of 1) at $z_0$, then show $f_1f_2$ has a pole of degree 2 at $z_0$ and provide an expression for $Res(f_1f_2; z_0)$. I.e, the residue of $f_1f_2$ at $z_0$.
This problem is trivial for rational functions, but I'm unsure how to approach this in general. Idea's I've had are to use the Laurent expansions for $f_1, f_2$, but I want to avoid this if possible. The idea is pretty simple, I just don't know of a short, concise way of proving it in general. Other ideas involve showing absolute convergence of the product of the Taylor series expansions of the denominators to rewrite the denominator as a single series where $z_0$ is a pole of degree 2. Not sure if that'd work though..
Thanks for any help!
 A: I know you said you wanted to avoid Laurent series, but it really isn't too painful.  I'll expand on Andrew's comment for the possibility that seeing more details will help you out.
Since $f_1$ and $f_2$ have simple poles at $z=z_0$ they can be written as Laurent series,
$$
f_1(z) = \sum_{k=-1}^{\infty} a_k (z-z_0)^k,
$$
$$
f_2(z) = \sum_{k=-1}^{\infty} b_k (z-z_0)^k,
$$
which converge in some annulus $0 < |z-z_0| < R$.  Multiplying these series term-by-term gives another Laurent series which converges in this annulus,
$$
f_1(z)f_2(z) = \sum_{k=-2}^{\infty} \left(\sum_{p+q = k} a_p b_q \right) (z-z_0)^k,
$$
which shows that $f_1f_2$ has a pole of order 2 at $z=z_0$.  This pole has residue 
$$
\sum_{p+q = -1} a_p b_q = a_{-1}b_0 + a_0b_{-1},
$$
which is the coefficient of $(z-z_0)^{-1}$ in this Laurent series.
(Note that the inner sum is the Cauchy product of the relevant coefficients of the individual Laurent series.)
If you'd like, these coefficients $a_{-1},a_0,b_{-1},b_0$ can be written down explicitly in terms of $f_1$ and $f_2$.  For example, it follows from the Laurent expansion of $f_1$ that
$$
a_{-1} = \lim_{z \to z_0} f_1(z)(z-z_0)
$$
and
$$
a_0 = \lim_{z \to z_0} \left(f_1(z) - \frac{a_{-1}}{z-z_0}\right).
$$
