Exercise · ZW-LAURENT-SERIES-003 · ★☆☆ Basic
On \(0<|z|<1\), how many terms with a negative power does the Laurent series of \(\frac1{z^2(1-z)}\) have?
Choose one option.
Show answer
- C. \(2\)
Explanation
With \(\frac1{1-z}=\sum_{k\ge0}z^k\) we get \(z^{-2}+z^{-1}+1+z+\cdots\): only \(z^{-2}\) and \(z^{-1}\) are negative.
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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