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Singular points and reidues

  1. Mar 20, 2010 #1
    I need to find and classify the singular points and find the residue at each of these points for the following function;

    f(z) = [tex]\frac{z^{1/2}}{z^{2}+1}[/tex]

    I can see that the singular points are at z=i and z=-i but have no idea how to classify them or find the residue at each point.

    I know finding the residue depends on if the singular points are removable singulairties, poles or zeros of certain orders but don't know to classify z=i or z=-i.

    Any help would be brillant, thankyou.
  2. jcsd
  3. Mar 20, 2010 #2


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    A function has a "pole of order n" at z= a if it can be written as a power series in z- a, the "Laurent series", including negative powers down to [itex](z-a)^{-n}[/itex].
    At an "essential singularity", such a power series includes all negative powers.
    If there is a pole at z= a, the residue at z= a is the coefficient of [itex](z-a)^n[/itex] in
    the powers series. We can write the example you give as [itex]\frac{z^{1/2}}{z+ i}\frac{1}{z- i}[/itex]. Since [itex]\frac{z^{1/2}}{z+i}[/itex] is analytic at z= i, it can be written as a power series in z- i with non-negative powers of z (Taylor's series)- which is of course, just it value at z= i. Multiplying that by [itex](z-i)^{-1}[/itex] gives a Laurent series down to (z- i)^{-1} so this function has a "pole of order 1" at z= i and its residue there is the same as the constant term of theTaylor's series for [itex]\frac{z^{1/2}}{z+ i}[/itex] around z= i which, after multplying by [itex]\frac{1}{z- i}= (z- i)^{-1}[/itex] gives the coefficient of [itex](z- i)^{-1}[/itex].

    In other words because [itex]z^{1/2}[/itex] is analytic at z= i and z= -i, [itex]\frac{z^{1/2}}{z^2+ 1}[/itex] has a pole of order 1 at both points. The residue at z= i is [itex]\frac{i^{1/2}}{i+ i}= \frac{i^{1/2}}{2i}[/itex] and the residuce at z= -i is [itex]\frac{(-i)^{1/2}}{-i- i}= -\frac{(-i)^{1/2}}{2i}[/itex].
    Last edited by a moderator: Mar 20, 2010
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