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Help figuring out this trigometric substitution

  1. Nov 29, 2009 #1
    From my notes I have

    w=u(x+iy)*(x^2 - y^2 +k^2 + i(2xy))^-.5

    We let N=x^2-y^2+k^2
    M=2xy
    R^2=(N^2+M^2)^2
    theta=tan^-1(M/N)

    using this, now

    w=u(x+iy)*(cos(theta/2)-isin(theta/2))*(x^2 - y^2 +k^2 )^2 + (2xy)^2 )^-.25

    I don't get that part. Btw, it simplifies to
    w=u(x+iy)*(cos(theta/2)-isin(theta/2)*(R^2)^-.25
     
  2. jcsd
  3. Nov 29, 2009 #2

    HallsofIvy

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    Science Advisor

    This has nothing to do with "Abstract and Linear Algebra" so I am moving it to "General Math".
     
  4. Nov 30, 2009 #3
    I haven't thought about that particular problem, but note that
    [tex](x+iy)^2=x^2-y^2+2ixy[/tex]

    So basically your equation is
    [tex]\frac{u(z)}{\sqrt{z^2+k^2}}[/tex]

    It seems you used [itex]z^2+k^2=Re^{i\theta}[/tex] hence [itex]R=\abs{z^2+k^2}[/itex] and [itex]\theta=\arg(z^2+k^2)[/itex]
     
    Last edited: Nov 30, 2009
  5. Nov 30, 2009 #4
    I figured it out. Let O=N+iM. Then let O=R*e^-i*theta
     
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