Exercise · ZW-LAURENT-SERIES-008 · ★★☆ Intermediate

Which integral gives the coefficient \(a_{-2}\) of the Laurent series of \(f\) about \(z_0\) (over a circle inside the annulus)?

Choose one option.

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  • A. \(\frac1{2\pi i}\oint f(\zeta)(\zeta-z_0)\,d\zeta\)

Explanation

The formula is \(a_n=\frac1{2\pi i}\oint\frac{f(\zeta)}{(\zeta-z_0)^{n+1}}d\zeta\). For \(n=-2\) the denominator exponent is \(-1\), i.e. we multiply by \((\zeta-z_0)\).

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Review the concept: Laurent series

The Laurent series of a function \(f\) holomorphic on an annulus \(r<|z-z_0|<R\) is its expansion in integer powers, positive and negative, of \(z-z_0\), with coefficients \(a_n\).

\[f(z)=\sum_{n=-\infty}^{\infty}a_n(z-z_0)^n\]

More practice problems: laurent series

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