<s>[INST] Compute $\oint_{|z|=2} \frac{d z}{1+z+z^{2}+z^{3}}$ with the residue theorem $$\oint_{|z|=2} \frac{d z}{1+z+z^{2}+z^{3}}$$
I calculated the above integral several times using the residue theorem.  The answer I get is $\pi / 2$ but this answer is not in the options please help

*

*$\frac{\pi i}{2}$

*$\frac{-\pi i}{2}$

*$\frac{\pi i}{1}$

*0

My answer
$\oint_{|z|=2} \frac{d z}{1+z+z^{2}+z^{3}}$
$1+z+z^{2}+z^{3}=0 \rightarrow z=\pm i, z=-1$
$\oint_{|z|=2} \frac{d z}{1+z+z^{2}+z^{3}}=2 \pi i\left(\left.\operatorname{Res}\left(\frac{1}{1+z+z^{2}+z^{3}}\right)\right|_{z=i}\right)$
$+\pi i\left(\left.\operatorname{Res}\left(\frac{1}{1+z+z^{2}+z^{3}}\right)\right|_{z=-1}\right)$
$=2 \pi i(-0.25-0.25 i)+0.5 \pi i=\frac{\pi}{2}$
 [/INST] You have:
$\oint_{|z|=2} \frac{d z}{1+z+z^{2}+z^{3}}$
$1+z+z^{2}+z^{3}=0 \rightarrow z=\pm i, z=-1$
After this you didn't apply the Residue Theorem correctly. Instead of
$\oint_{|z|=2} \frac{d z}{1+z+z^{2}+z^{3}}=2 \pi i\left(\left.\operatorname{Res}\left(\frac{1}{1+z+z^{2}+z^{3}}\right)\right|_{z=i}\right)+\pi i\left(\left.\operatorname{Res}\left(\frac{1}{1+z+z^{2}+z^{3}}\right)\right|_{z=-1}\right)$
It should be:
$\oint_{|z|=2} \frac{d z}{1+z+z^{2}+z^{3}}=2 \pi i\left(\left.\operatorname{Res}\left(\frac{1}{1+z+z^{2}+z^{3}}\right)\right|_{z=i}\right)\color{red}{+ 2 \pi i\left(\left.\operatorname{Res}\left(\frac{1}{1+z+z^{2}+z^{3}}\right)\right|_{z=-i}\right)}+\color{red}{2} \pi i\left(\left.\operatorname{Res}\left(\frac{1}{1+z+z^{2}+z^{3}}\right)\right|_{z=-1}\right)$
I don't know why you ignored the pole at $-i$, and you missed a $2$ in the term of the pole $-1$.
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