Problem

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Evaluate:

$$ \oint_{C}^{ }\left[\sum_{k=-\infty}^{\infty}\left(\frac{1}{z^{k}}+\frac{1}{z^{k-1}}\right)\right]dz $$

Where \(C\) is \(|z| = 1\)

Solution

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Split the sum into two parts:

$$ \oint_{C}^{ }\left[\sum_{k=-\infty}^{\infty}\frac{1}{z^{k}}+\sum_{k=-\infty}^{\infty}\frac{1}{z^{k-1}}\right]dz $$

Recall that the complex contour integral of all \(z^k\) is equal to zero, except in the case of \(\frac{1}{z}\), where it evaluates to \(2\pi i\). The left sum generates a \(\frac{1}{z}\) term at \(k=1\), and the right sum generates one at \(k=2\). Since there are two \(\frac{1}{z}\) terms, the entire integral evaluates to \(4\pi i\)