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Finding poles for cauchy's residue theorem.

  1. Feb 14, 2015 #1
    1. The problem statement, all variables and given/known data
    In order to use cauchy's residue theorem for a question, I need to put
    ##f(x)=\frac{z^{1/2}}{1+\sqrt{2}z+z^2}##
    Into the form
    ##f(x)=\frac{\phi(z)}{(z-z_0)^m}##.
    Where I can have multiple forms of
    ##{(z-z_0)^m}##
    on the denominator, e.g
    ##f(x)=\frac{z^{1/2}}{(z+1)(z+3)^3}##
    I just need to find what values of z will take it to zero
    2. Relevant equations


    3. The attempt at a solution
    The closest I have gotten is
    ##f(x)=\frac{Z^{1/2}}{(z+\frac{1}{\sqrt{2}})^2+\frac{1}{2}}##.
    But I need to get rid of that half on the end of the denominator (I think) in order to get to the form I want.
    ##f(x)=\frac{Z^{1/2}}{(z+\frac{\sqrt{2}}{2}+i)(z+\frac{\sqrt{2}}{2}-i)-\frac{1}{2}}##.
    was another close attempt. Can anyone help me find right factorisation?
     
  2. jcsd
  3. Feb 14, 2015 #2

    gneill

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    Staff: Mentor

    Use the quadratic formula on the denominator to extract its roots.
     
  4. Feb 14, 2015 #3

    vela

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    Staff Emeritus
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    From where you got, you can do this:
    $$\left(z + \frac{1}{\sqrt 2}\right)^2+\frac 12 = \left(z + \frac{1}{\sqrt 2}\right)^2- \left(\frac 1{\sqrt 2}i\right)^2$$ and then factor the difference of squares the usual way.
     
  5. Feb 15, 2015 #4
    Awesome this was exactly what I needed. THANKS!
     
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