1. ## inverse transformation?

Hi

I'm working on some questions and I'm just unsure how one of the solutions work.

I'm given

x=1/2(u^2-v^2)
y=uv
z=z

and in the solution it is transformed into

u=[x+(x^2+y^2)^1/2]^1/2
v=[-x+(x^2+y^2)^1/2]^1/2
z=z

I'm unsure how it was transformed. The question says to use "inverse transformation" but I'm not sure what that is.

Thanks!

2. Originally Posted by romtik
Hi

I'm working on some questions and I'm just unsure how one of the solutions work.

I'm given

x=1/2(u^2-v^2)
y=uv
z=z

and in the solution it is transformed into

u=[x+(x^2+y^2)^1/2]^1/2
v=[-x+(x^2+y^2)^1/2]^1/2
z=z

I'm unsure how it was transformed. The question says to use "inverse transformation" but I'm not sure what that is.

Thanks!
$u^2 - v^2 = 2x$ .... (1)

$uv = y \Rightarrow u = \frac{y}{v}$ .... (2)

Substitute (2) into (1):

$\left( \frac{y}{v} \right)^2 - v^2 = 2x$

$\Rightarrow y^2 - v^4 = 2x v^2$

$\Rightarrow v^4 + 2x v^2 - y^2 = 0$

whch is a quadratic in $v^2$. From the quadratic formula:

$\Rightarrow v^2 = \frac{-2x \pm \sqrt{4x^2 + 4y^2}}{2} = -x \pm \sqrt{x^2 + y^2}$

Discard the negative root solution since $v^2 > 0$:

$\Rightarrow v^2 = -x + \sqrt{x^2 + y^2}$

etc.