Exercise · ZW-ZERO-AND-POLE-007 · ★★☆ Intermediate

What is the order of the zero of \(e^z-1-z\) at \(z=0\)?

Choose one option.

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  • A. \(2\)

Explanation

Since \(e^z-1-z=\frac{z^2}{2}+\frac{z^3}{6}+\cdots\), the constant and linear terms vanish and the first nonzero one is \(z^2\): 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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