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
$\oint_{|z|=2} \frac{d z}{1+z+z^{2}+z^{3}}$
$1+z+z^{2}+z^{3}=0 \rightarrow z=\pm i, z=-1$
$\oint_{|z|=2} \frac{d z}{1+z+z^{2}+z^{3}}=2 \pi i\left(\left.\operatorname{Res}\left(\frac{1}{1+z+z^{2}+z^{3}}\right)\right|_{z=i}\right)$
$+\pi i\left(\left.\operatorname{Res}\left(\frac{1}{1+z+z^{2}+z^{3}}\right)\right|_{z=-1}\right)$
$=2 \pi i(-0.25-0.25 i)+0.5 \pi i=\frac{\pi}{2}$
 A: 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 $-1$.
A: $$f(z)=\frac{1}{1+z+z^{2}+z^{3}}=\frac{z-1}{z^{4}-1}$$
Then $-1,i,-i$ are simple poles.
When $f(z)=\frac{g(z)}{h(z)}$ and $h(z)$ has zero of order 1 at $z_{0}$ and $g(z_{0})\neq 0$. Then the residue is $\frac{g(z_{0})}{h'(z_{0})}$.
The residue is given by $\displaystyle Res_{z=z_{0}}\frac{z_{0}-1}{4z_{0}^{3}}$.
$$Res_{z=-1}f(z)=\frac{1}{2}$$
$$Res_{z=-i}f(z)=\frac{-1+i}{4}$$
$$Res_{z=i}f(z)=\frac{-1-i}{4}$$.
Add them to get $0$. And hence the integral is $0$ by Residue Theorem.
A: The function $$\frac{1}{z^3+z^2+z+1}$$ is holomorphic on the region outside $|z|=2$ on the Riemann sphere, i.e. $\hat{\mathbb{C}}\setminus B(0,2)$.
But that region is simply connected.  By Cauchy's theorem, we can contract the contour to a point within that region, and the integral of a point vanishes.
Thus the integral is $0$.
