will we still get full marks for this?
Say for example, HSC 1998, Q2 (e) (i)
By solving the equation z^3 + 1 = 0, fnd the three cube roots of -1.
If I do it by solving
z^3 = -1, -1 = cis(pi),
using De Moivre's theorem,
will that be ok?
I think you're meant to solve (z+1)(z^2 - z + 1) = 0
Say for example, HSC 1998, Q2 (e) (i)
By solving the equation z^3 + 1 = 0, fnd the three cube roots of -1.
If I do it by solving
z^3 = -1, -1 = cis(pi),
using De Moivre's theorem,
will that be ok?
I think you're meant to solve (z+1)(z^2 - z + 1) = 0