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Cool!
z⁴ + z² + 1 = z⁴ + 2z² + 1 - z² = (z² + 1)² - z² = (z² + z +1)(z² - z + 1) etc.
You could also factor z^4+z^2+1 into (z^2+z+1)(z^2-z+1).
Oh that’s right
The solutions are the complex cube roots of 1 and -1.
z(z^4+z^2+1)=0 therefore z=0 is one solution and z^4+z+1=0 means z^2 is a nonreal cubic root of unity, or better, z is a nonreal sixth rooth of unity, cos(kpi/3)+isin(kpi/3) for k=1,2,4,5
im wondering for the z^6=1 , why did you write 1 as e^2πmi and not e^i(2π + 2πm)
They are the same if you think about it
Cool!
z⁴ + z² + 1 = z⁴ + 2z² + 1 - z² = (z² + 1)² - z² = (z² + z +1)(z² - z + 1) etc.
You could also factor z^4+z^2+1 into (z^2+z+1)(z^2-z+1).
Oh that’s right
The solutions are the complex cube roots of 1 and -1.
z(z^4+z^2+1)=0 therefore z=0 is one solution and z^4+z+1=0 means z^2 is a nonreal cubic root of unity, or better, z is a nonreal sixth rooth of unity, cos(kpi/3)+isin(kpi/3) for k=1,2,4,5
im wondering for the z^6=1 , why did you write 1 as e^2πmi and not e^i(2π + 2πm)
They are the same if you think about it