Find $$\int_C \frac{1}{(z^4-1)}dz$$ where $C$ is the circle $|z+i|=\sqrt 3$.
So i tried to break the dominator, $\int_C \frac{1}{(z-1)(z+1)(z+i)(z-i)}dz$ and I knew that the domain is circle with center at $(-i,0)$ and $r=\sqrt 3$ so $z+i$ is point of singularity and $z-i$ is located outside the domain, but how can i know that $z=1$ or $z=-1$ is inside or outside the domain?? though the answer is $z=1$ and $z=-1$ is inside domain. and how can I made this into $|z-1|=\frac{3}{2}-\sqrt 2$, $|z+1|=\frac{3}{2}-\sqrt 2$ ? Thanks!!
A pole $z_k$ is inside the circle $C$ if and only if its distance from the center of the circle $-i$ is less then the radius $\sqrt{3}$, i.e. $$|z_k+i|^2<3.$$ Hence \begin{align} &z_1=1\implies |z_1+i|^2=|1+i|^2=1+1=2<3\implies \text{$1$ is inside $C$,}\\ &z_2=-1\implies |z_2+i|^2=|-1+i|^2=1+1=2<3\implies \text{$-1$ is inside $C$,}\\ &z_3=-i\implies |z_3+i|^2=|-i+i|^2=0<3\implies \text{$-i$ is inside $C$,}\\ &z_4=i\implies |z_4+i|^2=|i+i|^2=4>3\implies \text{$i$ is NOT inside $C$.} \end{align} Hence, in order to apply the Cauchy Residue Theorem, you have to evaluate the residue at $1$, $-1$, and $-i$.