Exercise · ZW-COMPLEX-TAYLOR-SERIES-006 · ★★☆ Intermediate

The coefficient of \((z-1)^3\) in the Taylor series of \(1/z\) at \(z_0=1\) is:

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  • C. \(-1\)

Explanation

\(\tfrac1z=\tfrac1{1+(z-1)}=\sum(-1)^n(z-1)^n\), so \(a_3=-1\).

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Review the concept: Complex Taylor series

The Taylor series of a function \(f\) holomorphic on a disc \(|z-z_0|<R\) is the power series in \(z-z_0\), centred at \(z_0\), whose coefficients \(a_n\) come from the derivatives of \(f\) at \(z_0\).

\[f(z)=\sum_{n=0}^{\infty}a_n(z-z_0)^n,\quad a_n=\frac{f^{(n)}(z_0)}{n!}\]

More practice problems: complex taylor series

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