Find a couple of complex numbers, such that one number is the square of the other and vice versa.

Rem: There will be more than one solution.

We want x=y^2 and y=x^2 thus x = x^4
One solution is x=0
others arise if
x³ = 1

In polar form
x = r(cos(θ)+isin(θ))
x³ = r³(cos(3θ)isin(3θ)) = 1

which requires r³=1 and 3θ=360º*n (n an integer)
so r=1
and θ can be one of 3 values
if n=0, θ=0º
if n=1, θ=120º
if n=2, θ=240º
(if n is over 2, these repeat)

Solutions:
(0,0) (1,1)  ((cos(120)+isin(120)), (cos(240)+isin(240))

Posted by Jer on 2014-11-19 10:40:05