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
x-y=y^2-x^2
-(y-x)=(y+x)(y-x)
first we have y-x=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 y-x is not zero then we have
y+x=-1
y=-x-1 giving
-x-1=x^2
x^2+x+1=0
x=(-1+-sqrt(-3))/2
x=-1/2+i*sqrt(3)/2 or x=-1/2-i*sqrt(3)/2
giving
y=-1/2-i*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,-1/2-i*sqrt(3)/2)
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Posted by Daniel
on 2014-11-19 08:28:56 |
solution
