Why doesn't an infinitely large term contribute to the residue of this double pole? Consider the function:
$$
f(z) = \frac{1}{(z+i)^2(z-i)^2}
$$
This function has two double-poles at $z_\pm = \pm i$.
Let's say that I wanted to find $\mathrm{Res}[f, z_+]$. One way of doing this is to make a change of variables $\xi = z - i$, and then to do a Taylor series around $\xi = 0$.
Doing this, we see:
$$
\begin{align}
f(z) &= \frac{1}{(\xi + 2i)^2\xi^2} \\
     &= \frac{1}{\xi^2}\cdot\frac{1}{(2i)^2}\cdot\frac{1}{\left(1 + \frac{\xi}{2i}\right)^2} \\
     &= -\frac{1}{\xi^2}\cdot\frac{1}{4}\left(1-\frac{\xi}{i} + O(\xi^2)\right) \\
     &= -\frac{1}{\xi}\cdot\frac{1}{4}\left(\frac{1}{\xi}-\frac{1}{i} + O(\xi)\right) 
\end{align}
$$
Now, the residue for this pole, I am told, is whatever the factor of $1/\xi$ is. In this case, we ignore the $O(\xi)$ terms since they go to zero, and thus find:
$$
\begin{align}
\mathrm{Res}[f, z_+] &= -\frac{1}{4\xi} + \frac{1}{4i} \\
                     &\rightarrow -\infty\ ?
\end{align}
$$
I claim this goes to infinity since $\xi \rightarrow 0$.
However, I am reliably informed that this term is to be ignored, and
$$
\mathrm{Res}[f, z_+] = \frac{1}{4i}
$$
instead. Why is this?
 A: The residue of $f(z)$ at $a$ is, by definition, the coefficient of $(z-a)^{-1}$ in the Laurent series centered at $a$.
To see why the coefficient of the $(z-a)^{-1}$ term is special, let's look at $a = 0$ and the function $f(z) = c z^{-n}$, where $n \in \mathbb{Z}$.  If we take the contour integral around a unit circle centered at $z = a = 0$, we have $z = e^{i \theta}$ and so
$$
\oint f(z) \, dz = \oint \frac{c}{e^{in\theta}} d(e^{i \theta}) = ic \int_0^{2 \pi} e^{-i(n-1)\theta} \, d \theta \\ = i c \int_0^{2\pi} \left[ \cos (n-1) \theta - i \sin (n-1) \theta \right] \, d \theta
$$
If $n \neq 1$, this integral will vanish, since we are integrating sinusoidal functions over an integer number of their periods.  Only if $n = 1$ does the integral turn out to be non-zero;  in this case, we have
$$
\oint f(z) \, dz = i c \int_0^{2\pi} d \theta = 2 \pi i c 
$$
In other words, the contour integral of $f(z)$ around this pole is
$$
\oint f(z) \, dz = \begin{cases} 2 \pi i c & n = 1 \\ 0 & n \neq 1 \end{cases}.
$$
The value $c$ is, of course, the residue of $f(z)$ at $z = 0$.
By extension, this means that if you have an integral that can be written as the sum of multiple terms like this—for example,
$$
f(z) = - \frac{1}{z^2} + \frac{1}{4 i z} + 3 z,
$$
then the contributions to the contour integral from the first and last terms are zero, and only the contribution from the middle term (the $z^{-1}$ term) contributes to the integral.
A: The residue is always a complex number.
In your case, is $\xi=z-i$, then\begin{align}f(z)&=f(\xi+i)\\&=\frac1{\xi ^4+4 i \xi ^3-4 \xi ^2}\\&=\frac1{\xi^2}\frac1{\xi^2+4i\xi^3-4}\\&=\frac1{\xi^2}\left(-\frac{1}{4}-\frac{i \xi }{4}+\frac{3 \xi^2}{16}+\frac{i \xi ^3}{8}+\cdots\right)\\&=-\frac{1}{4\chi^2}-\frac i{4\xi}+\frac{9}{16}+\frac{i \xi}8,\end{align}and therefore the residue is $-\frac i4$.
