Exercise · ZW-ZERO-AND-POLE-008 · ★★☆ Intermediate

What is the order of the pole of \(\frac{1}{z\,(e^z-1)}\) at \(z=0\)?

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  • D. \(2\)

Explanation

Both \(z\) and \(e^z-1=z+\frac{z^2}{2}+\cdots\) have a simple zero at \(0\); their product has a zero of order \(2\), so the reciprocal has a pole of order \(2\).

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Review the concept: Zero and pole

A zero of order \(m\) of a holomorphic function \(f\) is a point \(z_0\) where \(f\) vanishes with multiplicity \(m\), with holomorphic quotient \(g\); a pole of order \(m\) is a point where \(1/f\) has such a zero.

\[f(z)=(z-z_0)^m\,g(z),\qquad g(z_0)\neq 0\]

More practice problems: zero and pole

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