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.
x=y^2 (1)
y=x^2 (2)
(1)(2) gives
xy=y^2x^2
(yx)=(y+x)(yx)
first we have yx=0 as a solution
giving y=x thus
x=x^2 thus
x=0 or x=1 giving the solutions (0,0) and (1,1)
if yx is not zero then we have
y+x=1
y=x1 giving
x1=x^2
x^2+x+1=0
x=(1+sqrt(3))/2
x=1/2+i*sqrt(3)/2 or x=1/2i*sqrt(3)/2
giving
y=1/2i*sqrt(3)/2 and y=1/2+i*sqrt(3)/2 respectively
thus we have three distinct solutions
(0,0)
(1,1)
(1/2+i*sqrt(3)/2,1/2i*sqrt(3)/2)

Posted by Daniel
on 20141119 08:28:56 