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9.3: Behavior of functions near zeros and poles

  • Page ID
    6524
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    The basic idea is that near a zero of order \(n\), a function behaves like \((z - z_0)^n\) and near a pole of order \(n\), a function behaves like \(1/(z - z_0)^n\). The following make this a little more precise.

    Behavior near a zero. If \(f\) has a pole of order \(n\) at \(z_0\) then near \(z_0\),

    \[f(z) \approx \dfrac{b_n}{(z - z_0)^n}, \nonumber \]

    for some constant \(b_n\).

    \(Proof\). This is nearly identical to the previous argument. By definition \(f\) has a Laurent series around \(z_0\) of the form

    \[\begin{array} {rcl} {f(z)} & = & {\dfrac{b_n}{(z - z_0)^n} + \dfrac{b_{n - 1}}{(z - z_0)^{n - 1}} + \ ... \ + \dfrac{b_1}{z - z_0} + a_0 + \ ...} \\ {} & = & {\dfrac{b_n}{(z - z_0)^n} \left(1 + \dfrac{b_{n - 1}}{b_n} (z - z_0) + \dfrac{b_{n - 2}}{b_n} (z - z_0)^2 + \ ... \right)} \end{array} \nonumber \]

    Since the second factor equals 1 at \(z_0\), the claim follows.

    Picard’s Theorem and Essential Singularities

    Near an essential singularity we have Picard’s theorem. We won’t prove or make use of this theorem in 18.04. Still, we feel it is pretty enough to warrant showing to you.

    Theorem \(\PageIndex{1}\): Picard's Theorem

    If \(f(z)\) has an essential singularity at \(z_0\) then in every neighborhood of \(z_0\), \(f(z)\) takes on all possible values infinitely many times, with the possible exception of one value.

    Example \(\PageIndex{1}\)

    It is easy to see that in any neighborhood of \(z = 0\) the function \(w = e^{1/z}\) takes every value except \(w = 0\).

    Quotients of functions

    We have the following statement about quotients of functions. We could make similar statements if one or both functions has a pole instead of a zero.

    Theorem \(\PageIndex{2}\)

    Suppose \(f\) has a zero of order \(m\) at \(z_0\) and \(g\) has a zero of order \(n\) at \(z_0\). Let

    \[h(z) = \dfrac{f(z)}{g(z)}. \nonumber \]

    Then

    • If \(n > m\) then \(h(z)\) has a pole of order \(n - m\) at \(z_0\).
    • If \(n < m\) then \(h(z)\) has a zero of order \(m - n\) at \(z_0\).
    • If \(n = m\) then \(h(z)\) is analytic and nonzero at \(z_0\).

    We can paraphrase this as \(h(z)\) has ‘pole’ of order \(n - m\) at \(z_0\). If \(n - m\) is negative then the ‘pole’ is actually a zero.

    Proof

    You should be able to supply the proof. It is nearly identical to the proofs above: express \(f\) and \(g\) as Taylor series and take the quotient.

    Example \(\PageIndex{2}\)

    Let

    \[h(z) = \dfrac{\sin (z)}{z^2}. \nonumber \]

    We know \(\sin (z)\) has a zero of order 1 at \(z = 0\) and \(z^2\) has a zero of order 2. So, \(h(z)\) has a pole of order 1 at \(z = 0\). Of course, we can see this easily using Taylor series

    \[h(z) = \dfrac{1}{z^2} \left(z - \dfrac{z^3}{3!} + \ ... \right) \nonumber \]


    This page titled 9.3: Behavior of functions near zeros and poles 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; a detailed edit history is available upon request.