My professor gave us the the following integral to solve $$\oint_{C}\frac{z^{15}}{(z^2+1)^3(z^4-2)^2}dz \ \ , \ \ \vert{z}\vert=2 \ \ , \ \ \text{0 is the centre}$$ I know that the poles are $z=\pm{i}$ (m=$3$) for $(z^2+1)^3$ and the 4 simple poles that are the roots of $(z^4-2)$. My question is if i have to find all the residues and calculate it or if the integral is $0$ if the function is analytic.
2026-04-08 20:52:44.1775681564
Closed Contour Integration
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