Exercise · ZW-COMPLEX-TAYLOR-SERIES-002 · ★☆☆ Basic

The coefficient of \(z^5\) in the Taylor series of \(\sin z\) at \(0\) is:

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  • D. \(\tfrac1{120}\)

Explanation

\(\sin z=z-\tfrac{z^3}{3!}+\tfrac{z^5}{5!}-\cdots\), so \(a_5=1/120\).

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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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