Compute the integral $\int_{\delta_{B_1(-2)}} \frac{1}{z(z+2)^3}dz$ Could you please help me to solve this integral. I have no idea on how to proceed. Thank you.
I think, i should rewrite $\frac{1}{z(z+2)^3}$ as  a sum from a geometric series and then use the Cauchy integral formula, but I don´t see a trick..
$\int_{\delta_{B_1(-2)}} \frac{1}{z(z+2)^3}dz=\oint\frac{\frac{1}{z}}{(z+2)^3}dz$
Then
$\oint\frac{\frac{1}{z}}{(z+2)^3}dz=\left(\frac{2\pi i }{2!}\right)\left[\frac{d^2f(z)}{dz^2}\right]_{z=-2}$
$=\left(\frac{2\pi i }{2!}\right)\left[\frac{d^2}{dz^2}(\frac{1}{z})\right]_{z=-2}$
$=\left({\pi i}\right)\left[{\frac{2}{(z)^3}}\right]_{z=-2}$
$=\color{blue}{-\frac{1\pi i}{4}}$
Is that correct?



*Solution (geometric progression)

$\int_{\delta_{B_1(-2)}} \frac{1}{z(z+2)^3}dz$
$\frac{1}{z(z+2)^3} = 2*(1-(1+\frac{z}{2}))^{-1}*\frac{1}{z(z+2)^3}$
$=\frac{2}{z(z+2)^3}*\frac{1}{1-(1+\frac{z}{2})}$

We know, that
$\frac{1}{1-(1+\frac{z}{2})}=\sum_{n=0}^∞ {(1+\frac{z}{2})^n}=1+(1+\frac{z}{2})^1+(1+\frac{z}{2})^2+(1+\frac{z}{2})^3+ ...)$

So we get
$\frac{2}{z(z+2)^3}*\frac{1}{1-(1+\frac{z}{2})}=\frac{2}{z(z+2)^3}*(1+(1+\frac{z}{2})^1+(1+\frac{z}{2})^2+(1+\frac{z}{2})^3+ ...)=$
$=\frac{4}{(z+2)^3}+\frac{z}{(z+2)^3}+\frac{1/2}{(z+2)}+\frac{1}{4}+\frac{2+z}{8}+...$
=>
$f(z)=1/2, z=-2
=>
2\pi if(z)=2\pi i*1/2=\pi i$
In this solution I have $\color{blue}{\pi i}$
And in the top solution I have another answer $\color{blue}{-\frac{\pi i}{4}}$.
What's my mistake?

The only enclosed pole is a third-order one at $z=-2$, so the residue is $\tfrac12\lim_{z\to-2}\tfrac{d^2}{dz^2}[(z-2)^3f(z)]$. I leave you to evaluate this, then multiply by $2i\pi$.
