Relativistic kinetic energy of rigid bodies remix

x^2 + y^2 = c^2

x=a+b*i, y=u+v*i, c=n+m*i

(a+b*i)^2 + (u+v*i)^2 = (n+m*i)^2

(a^2 + abi + abi - b^2) + (u^2 + uvi + uvi -v^2) = n^2 + nmi + nmi - m^2

a^2 - b^2 + u^2 - v^2 - n^2 + m^2 = 2nmi - 2abi - 2uvi

a^2 - b^2 + u^2 - v^2 - n^2 + m^2 = 2i(nm - ab - uv)

a^2 - b^2 + u^2 - v^2 - n^2 + m^2
___________________________________ = i
                2(nm - ab - uv)



There are three cases of special relativity that remain untreated, first, when the

square root itself is negative, secondaly when the number beneath the root is imaginary.

Thirdly, when the root is negative and the lorentz factor imaginary.



First imaginary numbers result when (v^2/c^2) > 1

Using Relativistic_kinetic_energy_of_rigid_bodies from

en.wikipedia.org/wiki/Kinetic_…

M(1-v^2/c^2)^-1/2*v^2*i + Mc^2(1-v^2/c^2)^1/2*i ± E0  (the constant of integration can be negative)

Where M is mass, c is speed of light. It simplifies to

M(1-v^2/c^2)^-1/2*c^2*i ± E0     When (v^2/c^2) > 1

And when (1-v^2/c^2)^-1/2 is negative

Since "Every positive number a has two square roots: "(a)^1/2, which is positive, and -(a)^1/2, which is negative."

en.wikipedia.org/wiki/Square_r…

-M(1-v^2/c^2)^-1/2*c^2 ± E0 = kinetic energy

And the situation when the number is both imaginary and the root negative.

-M(1-v^2/c^2)^-1/2*c^2*i ± E0 = kinetic energy

And using the pythagore, vector combo theorem from above, we get real numbers again.

                                         a^2 - b^2 + u^2 - v^2 - n^2 + m^2
±M(1-v^2/c^2)^-1/2*c^2 *___________________________________   ± E0 = kinetic energy
                                                     2(nm - ab - uv)