Let
$$f(z)=\frac{e^{-z}}{z(\cos{z}-1)}$$
a) Determine the order of $f$:s pole at $z=0$.
b) Find the four first terms $\neq 0$ in $f$:s Laurent expansion valid in the region $\{z:0<|z|<2\pi\}.$
c) Evaluate the integral
$$\int_{|z|=1}f(z) \ dz$$
As soon as I've Laurent expanded $f(z)$ around $z=0$, the answers to a), b) and c) are trivial.
I have that
$$e^{-z}=1-z+\frac{z^2}{2}-\frac{z^3}{6}+O(z^4),$$
$$z(\cos{z}-1)=-\frac{z^3}{2}+\frac{z^5}{24}+O(z^7).$$
EDIT:
$$\frac{1}{z(\cos{z}-1)}=\frac{1}{-\frac{z^3}{2}+\frac{z^5}{24}+O(z^7)}=-\frac{2}{z^3}\frac{1}{1-\frac{z^2}{12}+O(z^4)}=-\frac{2}{z^3}\left(1+\frac{z^2}{12}+O(z^4)\right)=\left(-\frac{2}{z^3}-\frac{1}{6z}+O(z)\right),$$
so
$$\frac{e^{-z}}{z(\cos{z}-1)}=\left(-\frac{2}{z^3}-\frac{1}{6z}+O(z)\right)\left(1-z+\frac{z^2}{2}-\frac{z^3}{6}+O(z^4)\right)$$
$$=-\frac{2}{z^3}+\frac{2}{z^2}-\frac{7}{6z}+\frac{1}{2}+O(z).$$
Now all the assignments follow immediately. $\text{Res}_{0}(f)=-7/6$ so the integral is equal to $-7\pi i/3$, by the residual theorem.
First of all, we may write $f$ as $$ f(z)=-\frac{e^{-z}(\cos(z)+1)}{z\sin^2(z)}. $$
My hint is to calculate the limits of the form: $$ \lim_{z\to z_0}\ (z-z_0)^{n}f(z), $$ for each $n\in\mathbb N$, with $z_0=0$. If the limit does not converge for any $n\in\mathbb N$, then $z_0$ is essential. If the limit converges to $0$ for some $n\in\mathbb N$, then we consider $n_0$ the first $n$ for which this happens. If $n>1$,then $z_0$ is a pole of order $n-1$. If $n=1$,then $z_0$ is a removable singularity of $f$.
Note that, for $n=3$, $$ \lim_{z\to z_0}\ (z-z_0)^3f(z) =\lim_{z\to0}\ -\frac{e^{-z}(\cos(z)+1)}{\left(\frac{\sin(z)}{z}\right)^2} = -2. $$ Then, for any $n>3$, $$ \lim_{z\to z_0}\ (z-z_0)^nf(z) =\left(\lim_{z\to0}\ z^{n-3}\right)\left(\lim_{z\to0}\ \frac{e^{-z}}{\frac{\cos(z)-1}{z^2}}\right) = 0. $$ That means that $0$ is a pole of $f$ of order $3$.
This means that the Laurent series of $f$ around $0$ is of the form $$ f(z)=\sum_{n=-3}^{+\infty}a_nz^n, $$ Then, putting $g(z)=z^3f(z)$, we have that $$ g(z)=\sum_{n=0}^{+\infty}a_{n-3}z^n, \forall 0<|z|<r, $$ for some $r>0$. Since the series has only positive exponents, we may also calculate it at the point $0$. Define $g$ at $0$ by $a_{-3}$. Since $g$ is given by an infinite polynomial which converges in any point of $D(0,r)$, it follows that $g$ is holomorphic in $D(0,r)$. Therefore, the derivative $g^{(j)}$ is continuous for every integer $j\geq0$ and the $k$-th coefficient of the expansion of $g$ equals to $\frac{1}{k!}g^{(j)}(0)$. Using these facts we get: $$ \begin{array}{rl} a_{-3} & =g(0)=\lim_{z\to0} g(z) = \lim_{z\to0}\ -\frac{e^{-z}(\cos(z)+1)}{\left(\frac{\sin(z)}{z}\right)^2} = -2; \\ a_{-2} & =g'(0)= \displaystyle\lim_{z\to0} g'(z) = \lim_{z\to0}\ \left(-\frac{e^{-z}(\cos(z)+1)}{\left(\frac{\sin(z)}{z}\right)^2}\right)' \\ & = \displaystyle\lim_{z\to0}\ -\frac{\left[-e^{-z}(\cos(z)+1)+e^{-z}(-\sin(z))\right]\left(\frac{\sin(z)}{z}\right)^2 - 2\frac{\sin(z)}{z}\left(\frac{z\cos(z)-\sin(z)}{z^2}\right)}{\left(\frac{\sin(z)}{z}\right)^4}, \end{array} $$ and using L'Hopital we see that $\displaystyle\lim_{z\to0}\frac{z\cos(z)-\sin(z)}{z^2}=0$, and consequently, $$ a_{-2} = e^{-0}(\cos(0)+1)+e^{-0}\sin(0) = 2; $$ To continue, we must evaluate: $$ \begin{array}{rl} a_{-1} & =g''(0)=\lim_{z\to0} g''(z) = \lim_{z\to0}\ \left(\frac{e^{-z}(\cos(z)+1)}{\left(\frac{\sin(z)}{z}\right)^2}\right)''; \\ a_{0} & =g^{(3)}(0)=\lim_{z\to0} g^{(3)}(z) = \lim_{z\to0}\ \left(\frac{e^{-z}(\cos(z)+1)}{\left(\frac{\sin(z)}{z}\right)^2}\right)^{(3)}. \end{array} $$
This is a way to solve this exercise. The calculations of the two last limits are very extense, I only calculated until the second one. I'm also very curious about a way to expand the Laurent series of this function directly. I hope it was helpful.