How to calculate the integral $\frac{1}{2\pi i}\oint_{|z|=1}\frac{2(1-\cos(z))e^z}{z^4}dz$ My try:
$$I = \frac{1}{2\pi i}\oint_{|z|=1}\frac{2(1-\cos(z))e^z}{z^4}dz$$
using the identity of Laurent's expansian $$a_{n} = \frac{1}{2\pi i}\oint_{C}\frac{f(z)}
{(z-z_{0})^{n+1}}dz.$$
I know for a fact that $z_0 = 0 $ and $n=3$, I also know now that
$$f(z) = 2(1-\cos(z))e^z .$$
I expanded this function into a series:
$$f(z) = 2\left(1-\sum(-1)^n\frac{z^{2n}}{(2n)!}\right) \cdot \sum \frac{z^n}{n!} = 2\sum (-1)^n\frac{z^{2n}}{(2n)!}+(-1)^{n+1}\cdot\frac{z^{2n}}{(2n)!}\cdot\frac{z^n}{n!}.$$
At $n=3$ and $z\to0$:
$$\lim_{n\rightarrow z_{0} = 0} \left(-\frac{z^{6}}{6!}+\frac{z^{6}}{6!}\cdot\frac{z^3}{3!}\right) = 0 .$$
I get wrong answer as the correct answer in my text book is 1 .
I would appreciate some help
 A: Cauchy's Integral formula immediately gives you that your integral is
$$
\frac1{3!}\,[2(1-\cos z)e^z]'''|_{z=0}=\frac16\,(2 e^z (1 + 2 \cos z + 2 \sin z))\Big|_{z=0}=\frac66=1.
$$
If alternatively you want to work with the expansions, you have
\begin{align}
\frac2{z^4}(1-\cos z)e^z&=\frac2{z^4}\Big(\frac{z^2}2+o(z^4)\Big)\Big(1+z+\frac{z^2}2+o(z^3)\Big)\\[0.3cm]
&=\frac2{z^4}\Big(\frac{z^2}2+\frac{z^3}2+o(z^4)\Big)\\[0.3cm]
&=\frac1{z^2}+\frac1z+o(1).
\end{align}
Now, we you integrate, the integral of $1/z^2$ is zero (via direct calculation or integral formula), the $1/z$ term integrates to $2\pi i$, and the $o(1)$ term is analytic and so integrates to zero. Thus the result is
$$
\frac{1}{2\pi i}\oint_{|z|=1}\frac{2(1-\cos(z))e^z}{z^4}dz=\frac{2\pi i}{2\pi i}=1. 
$$
A: We care about the coefficient of the $\frac {1}{z}$ term in the Laurent series.  This means we care about the coefficient of the $z^3$ in the numerator
$2(\frac{z^2}{2}+\cdots)(1+z+\cdots) = z^2 + z^3 + \cdots$
The rest of the expansions will create terms of higher degree than $z^3$
