Help figuring out this trigometric substitution

  • Thread starter ComFlu945
  • Start date
  • #1
9
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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
 

Answers and Replies

  • #2
HallsofIvy
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This has nothing to do with "Abstract and Linear Algebra" so I am moving it to "General Math".
 
  • #3
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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:
  • #4
9
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I figured it out. Let O=N+iM. Then let O=R*e^-i*theta
 

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