The value of the integral $\int_{C} \dfrac{z^2+1 dz}{(z+1)(z+2)}$ where $C$ is $|z|=\dfrac{3}{2}$ is: a) $0$ b) $\pi i$ c)$2\pi i$ d) $4\pi i$
I have tried:
Resolving them into factors by partial fraction method, which gave me: $\int_{C} [1+\dfrac{2}{z+1}-\dfrac{5}{z+2}]dz$ Here, I use the Cauchy Integral formula, given by $f^n(a)=\dfrac{n!}{2\pi i} \int_{C} \dfrac{f(z)dz}{(z-a)^{n+1}}$
with $n=0,a=-1$ for the second part and then $n=0,a=-2$ for the third part.
For the first part $\int_{C} 1 dz=0$ since $1$ is an entire function on C.
This leaves me with the final answer: $0+2\pi i- 2\pi i=0$. The answer given is $4\pi i$. Can you please point out my mistake and suggest any other method for doing this? I am new to complex integration and have trouble solving problems with the curve of the form $|z|=c$. Thank you!
Your function has two poles at $z=-1$ and $z=-2$. But only $-1$ is in the disk you want to consider. Hence, by Cauchy's formula $$\int_{C} \dfrac{z^2+1 dz}{(z+1)(z+2)}=2i\pi \left(\frac{z^2+1}{z+2}\right)|_{z=-1}=4i\pi.$$