Exercise · ZW-LAURENT-SERIES-009 · ★★★ Advanced
Find the coefficient of \((z-1)^{-1}\) in the Laurent series of \(\frac{e^{z}}{(z-1)^2}\) on \(0<|z-1|\).
Choose one option.
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- C. \(e\)
Explanation
With \(u=z-1\): \(\frac{e^{1+u}}{u^2}=e\left(u^{-2}+u^{-1}+\frac12+\cdots\right)\). The coefficient of \(u^{-1}\) is \(e\).
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 \(a_n\).
\[f(z)=\sum_{n=-\infty}^{\infty}a_n(z-z_0)^n\]
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