<s>[INST] Integral Of $\oint_{|z-1|=2} \frac{\cos \pi z}{(z^2-z)\cdot (z+2i)}dz$ I want to integrate the following :
$$\oint_{|z-1|=2} \frac{\cos \pi z}{(z^2-z)\cdot (z+2i)}dz$$
What I did:


*

*$$\oint_{|z-1|=2} \frac{\cos \pi z}{z+2i}\cdot \left [ \frac{1}{z(z-1)}\right]dz$$

*$$\left [ \frac{1}{z(z-1)}\right ] \rightarrow \frac{A}{z}+\frac{B}{z-1},A=-1,B=1$$

*$$\oint_{|z-1|=2} \frac{\cos \pi z}{z+2i}\cdot \left [ \frac{-1}{z}+\frac{1}{z-1}\right]dz$$



In this step I need to evaluate so:
$$\dots=-2\pi i\cdot f(0)+2\pi i \cdot f(1) = -2\pi i\cdot\frac{1}{2i}+2\pi i\cdot\frac{cos \pi}{1+2i}=?$$
Now what? I need to present it as Complex? this is the final answer? Any suggestions?
Thanks.
 [/INST] You may use the Cauchy's integral formula, http://en.wikipedia.org/wiki/Cauchy%27s_integral_formula.
$$
\oint_{|z-1|=2} \frac{\cos \pi z}{(z^2-z)\cdot (z+2i)}dz=I_1+I_2,
$$
where
$$
I_1=\oint_{|z-1|=2} \frac{\frac{\cos \pi z}{(z+2i)z}}{z-1}dz=2\pi i\frac{\cos\pi}{1+2i},
$$
and
$$
I_2=\oint_{|z-1|=2} \frac{\frac{\cos \pi z}{(z+2i)(z-1)}}{z}dz=2\pi i\frac{\cos0}{-2i.}
$$
$I_1+I_2=-\frac{\pi}{5} (9+2 i)$.
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