Exercise · ZW-RESIDUE-004 · ★★☆ Intermediate

What is the residue of \(\frac{1}{z^2+1}\) at \(z=i\)?

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  • C. \(-\frac{i}{2}\)

Explanation

It is \(\frac{1}{(z-i)(z+i)}\), a simple pole: the residue is \(\frac{1}{z+i}\) evaluated at \(i\), that is, \(\frac{1}{2i}=-\frac i2\).

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Review the concept: Residue

The residue of a function \(f\) at an isolated singularity \(z_0\) is the coefficient \(a_{-1}\) of \((z-z_0)^{-1}\) in its Laurent series on a punctured disc of radius \(r\).

\[\operatorname{Res}(f,z_0)=a_{-1}=\frac{1}{2\pi i}\oint_{|z-z_0|=r/2}f(z)\,dz\]

More practice problems: residue

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