Exercise · ZW-COMPLEX-TAYLOR-SERIES-008 · ★★★ Advanced

If \(f\) is entire, its Taylor series at \(0\) converges to \(f(z)\):

Choose one option.

Show answer
  • D. For every \(z\in\mathbb{C}\)

Explanation

There are no singularities at finite distance, so the radius of convergence is \(\infty\) and the series represents \(f\) on the whole plane.

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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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