Trigonometric substitution for complex square root

[ComFlu945]

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]

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

 

[Gerenuk]

I haven't thought about that particular problem, but note that
$$(x+iy)^2=x^2-y^2+2ixy$$

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

It seems you used [itex]z^2+k^2=Re^{i\theta}[/tex] hence $R=\abs{z^2+k^2}$ and $\theta=\arg(z^2+k^2)$[/itex]

 

[ComFlu945]

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