Calculating Residues around given poles using limit fomula i am having an issue with a sample question I have been given and I cannot figure out the correct way of dealing with the question.
I have to calculate the following residue of $\displaystyle\frac{\cos(z^3) - 4e^z}{z^8 - z}$ at $z = 0$
So, usually I would use the limit formula like this one: Linky
but I do not know what the order of the function is, i guess that is my main problem.  I know 7 is incorrect but how do I go about finding the order?
Thanks
 A: Let $f(z)=\frac{\cos(z^3)-4e^z}{z^8-z}$.  Note that the numerator approaches $-3$ as $z\to 0$.  Furthermore, note that $z^8-z=z(z^7-1)$ has a zero of order $1$.  Then, we deduce that $f$ has a pole of order $1$ and
$$\begin{align}
\text{Res}\left(\frac{\cos(z^3)-4e^z}{z^8-z},z=0\right)&=\lim_{z\to 0}\left(\frac{\cos(z^3)-4e^z}{z^7-1}\right)\\\\
&=3
\end{align}$$
A: Just expand the function at $z=0$ and use the definition of residue, i.e. IT IS the coefficient of $1/z$:
$$f(z):=\frac{\cos(z^3) - 4e^z}{z^8 - z}=\frac{1+O(z^6) - 4(1+z+O(z^2))}{-z(1-z^7)}
=\frac{3}{z}+O(1)$$
Therefore the residue at $z=0$ is $3$.
P.S. You can also compute the residue in using the linked formula: the pole $z=0$ has order $1$ therefore the residue is given by the limit
$$\lim_{z\to 0} zf(z)=\lim_{z\to 0}\frac{\cos(z^3) - 4e^z}{z^7 - 1}=\frac{1-4}{-1}=3.$$
A: Expanding the numerator around $z=0$ you get
$$ \cos(z^3) - 4 e^z = -3 - 4z + \mathcal O(z^6). $$
For the denominator you have instead
$$z(z^7-1) = -z + \mathcal O(z^8),$$
so that you can write the whole expression as
$$\frac{3}{z} + \mathcal O(1). $$
We conclude that the pole is of order 1, and the residue is 3.
