Exercise · ZW-COMPLEX-TAYLOR-SERIES-010 · ★★★ Advanced

The coefficient of \(z^3\) in the Taylor series of \(\dfrac{e^z}{1-z}\) at \(0\) is:

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  • C. \(\tfrac83\)

Explanation

It is the product of series: \(a_3=1+1+\tfrac12+\tfrac16=\tfrac83\) (sum of the coefficients of \(e^z\) up to \(z^3\)).

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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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