Exercise · ZW-COMPLEX-TAYLOR-SERIES-003 · ★☆☆ Basic

The radius of convergence of the Taylor series of \(\dfrac1{1-z}\) at \(0\) is:

Choose one option.

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  • A. \(1\)

Explanation

The series is \(\sum z^n\) and converges for \(|z|<1\); the pole \(z=1\) is at distance \(1\) from the centre.

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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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