The question is to evaluate the integral :
$$
\oint\limits_{|z-1|=2}\frac{\mathrm{d}z}{z^2(z^2-4)e^z}
$$
by using Cauchy's integral formula.
I have this so far:
We have a circle with radius 2 centered at 1, and we have 3 singularities, $z=2, z=-2, z=0$. However, $z=-2$ is not included within our circle, so we don't need to worry about it.
Now we can rewrite what we have as
$$
\oint\limits_{|z-1|=2}\frac{dz}{z^2(z^2-4)e^z} = \oint\limits_{|z|=\epsilon} \frac{\frac{e^{-z}}{z^2-4}}{z^2} \mathrm{d}z+
\oint\limits_{|z-2|=\epsilon}\frac{\frac{e^{-z}}{z^2(z+2)}}{z-2} \mathrm{d}z.
$$
My solution ends up as $\frac{i\pi}{2} $+ $\frac{i\pi}{8e^2}$.
I think this is correct, but when I do my calculations I end up with the wrong solution (the solution is in the back of the book). Can anyone guide me in the right direction, or tell me if what I'm doing is wrong? The answer given in the back of the book is given as $\frac{-i\pi}{2} +\frac{i\pi}{4e^2}.$
Best Answer
The pole $\;z=0\;$ has multiplicity $\;2,\;$ so $$2\pi i\underset0{\text{ Res }}\dfrac{e^{-z}}{z^2(z^2-4)} = 2\pi i\, \lim\limits_{z\to0}\left(\dfrac{e^{-z}}{z^2-4}\right)' = \lim\limits_{z\to 0}\,e^{-z} \left(\dfrac1{4-z}-\dfrac{2z}{(z^2-4)^2}\right)\\ = 2\pi i\cdot \dfrac14 = i\,\frac\pi2,$$ and this method applies quite correct way to get $\;c_{-1}\;$ from the Laurent series in the form of $$-\frac1{4 z^2} + \frac1{4 z} - \frac3{16} + \frac5{48}z+\dots$$
At the same time, the pole $\;z=2\;$ is a simple one, and $$2\pi i \underset{2}{\text{ Res }}\,\dfrac{e^{-z}}{z^2(z^2-4)} = 2\pi i\,\lim\limits_{z\to 2}\dfrac{e^{-z}}{z^2(z+2)} = \dfrac{2\pi i}{16e^2},$$ wherein the Laurent series also corresponds to the obtained result.