9.1: Poles and Zeros

We remind you of the following terminology: Suppose \(f(z)\) is analytic at \(z_0\) and \[f(z) = a_n (z - z_0)^n + a_ (z - z_0)^ + \ . \nonumber \] with \(a_n \ne 0\). Then we say \(f\) has a zero of order \(n\) at \(z_0\). If \(n = 1\) we say \(z_0\) is a simple zero. Suppose \(f\) has an isolated sigularity at \(z_0\) and Laurent series \[f(z) = \dfrac <(z - z_0)^n>+ \dfrac><(z - z_0)^> + \ . + \dfrac + a_0 + a_1 (z - z_0) + \ . \nonumber \] which converges on \(0 < |z - z_0| < R\) and with \(b_n \ne 0\). Then we say \(f\) has a pole of order \(n\) at \(z_0\). If \(n = 1\) we say \(z_0\) is a simple pole. There are several examples in the Topic 8 notes. Here is one more

Example \(\PageIndex\)

\[f(z) = \dfrac \nonumber \] has isolated singularities at \(z = 0\), \(\pm i\) and a zero at \(z = -1\). We will show that \(z = 0\) is a pole of order 3, \(z = \pm i\) are poles of order 1 and \(z = -1\) is a zero of order 1. The style of argument is the same in each case. At \(z = 0\): \[f(z) = \dfrac \cdot \dfrac. \nonumber \] Call the second factor \(g(z)\). Since \(g(z)\) is analytic at \(z = 0\) and \(g(0) = 1\), it has a Taylor series \[g(z) = \dfrac = 1 + a_1 z + a_2 z^2 + \ . \nonumber \] Therefore \[f(z) = \dfrac + \dfrac + \dfrac + \ . \nonumber \] This shows \(z = 0\) is a pole of order 3. At \(z = i\): \(f(z) = \dfrac \cdot \dfrac\). Call the second factor \(g(z)\). Since \(g(z)\) is analytic at \(z = i\), it has a Taylor series \[g(z) = \dfrac = a_0 + a_1 (z - i) + a_2 (z - i)^2 + \ . \nonumber \] where \(a_0 = g(i) \ne 0\). Therefore, \[f(z) = \dfrac + a_1 + a_2 (z - i) + \ . \nonumber \] This shows \(z = i\) is a pole of order 1. The arguments for \(z = -i\) and \(z = -1\) are similar.

This page titled 9.1: Poles and Zeros is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Jeremy Orloff (MIT OpenCourseWare) via source content that was edited to the style and standards of the LibreTexts platform.

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