Help figuring out this trigometric substitution

**ComFlu945** · Nov29-09, 08:11 PM

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

**HallsofIvy** · Nov29-09, 09:56 PM

This has nothing to do with "Abstract and Linear Algebra" so I am moving it to "General Math".

**Gerenuk** · Nov30-09, 10:56 AM

I haven't thought about that particular problem, but note that

So basically your equation is
$LaTeX Code: \\frac{u(z)}{\\sqrt{z^2+k^2}}$

It seems you used $LaTeX Code: z^2+k^2=Re^{i\\theta}$ hence $LaTeX Code: R=\\abs{z^2+k^2}$ and $LaTeX Code: \\theta=\\arg(z^2+k^2)$

**ComFlu945** · Nov30-09, 04:54 PM

I figured it out. Let O=N+iM. Then let O=R*e^-i*theta