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Order of zero (complex analysis)

  1. Feb 15, 2016 #1
    I'm confused about how this related to the roots of an equation. Here's the definition:

    We say that ##f## has a zero of order ##m## at ##z_0## if
    ##f^{(k)}(z_0)=0## for ##k=0,\dots , m-1##, but ##f^{(m)}(z_0)\ne 0##
    or equivalently that
    ##f(z) = \sum_{k=m}^\infty a_k(z-z_0)^k##, ##a_m \ne 0##.

    For example the function ##(z^2+z-2)^3## has zeros of order ##3## for ##z=2## and ##z=-1## which is the multiple of the roots so the easiest way to determine the order seems to just note this instead of differentiating ##4## times.

    However for a function like ##z^2(1-\cos z)## this doesn't work anymore. If you differentiate this you actually find a zero of order ##4## at ##z=0## and zeros of order ##2## at ##z = 2\pi n, \; = \pm 1, \pm 2, \dots## which isn't what I would expect just looking for roots to the equation.

    So when determining the order of a zero is the approach always to differentiate? Since for simple functions like polynomials they seem to be the same as the roots.
     
    Last edited: Feb 15, 2016
  2. jcsd
  3. Feb 15, 2016 #2

    RUber

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    I take it you mean for your example:
    ##f(z) = (z^2 + z -2) ^ 3## has zeros at ## z=-2, 1##.
    For polynomials, you are correct that the multiplicity of the repeated roots is the same as the order of the zero at those roots.
    Example:
    ##p(z)=(z-z_0)^m (z-z_1)^n ## has a root of multiplicity m at ## z=z_0## and a root of multiplicity n at ## z = z_1##.

    If you look at the derivatives up to the (m-1)th, you will notice that all the terms will include at least one power of ##(z-z_0)##.
    However, the mth derivative will have one term where the ##(z-z_0)## has been reduced to a constant times ##(z-z_1)^n ##.
    Assuming that ##z_0 \neq z_1##, this means that the order of the zero at ##z=z_0## is m.

    Therefore, you can show, in general, for all polynomials that the multiplicity of a root is the same as the order of the zero at that root.


    If you differentiate your example, you get:
    ## f^1(z) = 3(z^2 + z -2)^2 (2z + 1) ## which as zeros at ## z=-2, 1## and also ##z=-1/2##.
    And again,
    ## f^2(z) = 6(z^2 + z -2) (2z + 1)^2 + 6(z^2 + z -2)^2## which as zeros at ## z=-2, 1## and not ##z=-1/2##.
    Once more:
    ## f^3(z) = 6 (2z + 1)^3+24(z^2 + z -2) (2z + 1) + 12(z^2 + z -2)(2z+1)## has a zero at ##z=-1/2## and not ## z=-2, 1##
    So, using the derivative test, you confirm that -2 and 1 are zeros of order 3.
     
  4. Feb 15, 2016 #3
    Thanks! Writing it that way make it pretty obvious for polynomials.

    From your example (and my second one) it also seems we have a product rule, that is if ##f(z) = g(z)h(z)## and ##g(z)## has a zero at ##a_0## of order ##n## and ##h(z)## have a zero of order ##m## at ##a_0## then ##f(z)## have a zero of order ##n+m## at ##a_0##.

    This seem to follow from Leibniz rule , ##f^{(n+m-1)}=(gh)^{(n+m-1)} = \sum_{k=0}^{n+m-1} \binom{n+m-1}{k} g^{(k)}h^{(n+m-1-k)}## , since ##g^{(k)}= 0## for all ##k < n## while ##h^{(n+m-k)}=0## for ##n\le k \le (n+m-1)## hence every term is zero. Similar applies for every higher derivative in between ##0## and ##(n+m-1)##.
    While for the next order ##(n+m)## we have the only non-zero term ##f^{(n)}g^{(m)}## left.
     
  5. Feb 15, 2016 #4
    Never mind second half of my post, apparently there's a really obvious way to see this directly.
    From the theorem stating that:
    If ##f## has a zero of order ##m## at ##z_0##, then ##f(z) = (z-z_0)^mg(z)##, where ##g## is analytic in ##D## and ##g(z_0) \ne 0##. This makes it really obvious when multiplying two functions, and also the polynomial part.

    Anyway thanks again for the explanation, I believe I understand the order of zeros now!
     
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