I factorized $\frac{1}{z^4 - 1}$ into $\frac{1}{(z-1)(z+1)(z-i)(z+i)}$ and I know that it will be somehow solved by partial fraction but not able to proceed further. Answer for this question is showing $0$.
2026-03-29 20:20:44.1774815644
$\oint \frac{dz}{z^4 - 1}$ along $|z|=2$ using Cauchy integral formula
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Hint:
We have $$\frac{1}{(z-1)(z+1)(z-i)(z+i)}=\frac{1}{4(z-1)}-\frac{1}{4(z+1)}+\frac{i}{4(z-i)}-\frac{i}{4(z+i)}$$
Now use that $$\oint_{|z|=2}\frac{dz}{z-a}=2\pi i$$ where $|a|<2$.