Oh right I see, thank you!If w is a complex cube root of unit, then 1+w+w^2=0.
So we have 1-w+w^2 = -2w, and 1+w-w^2=-2w^2.
Hence, the LHS is (-2w)^3 - (-2w^2)^3 = -8w^3 + 8w^6 = -8+8 (since w^3=1) and we thus have 0.
Sorry, how did you get 1+w+w^2=0 just from w^3=1?If w is a complex cube root of unit, then 1+w+w^2=0.
So we have 1-w+w^2 = -2w, and 1+w-w^2=-2w^2.
Hence, the LHS is (-2w)^3 - (-2w^2)^3 = -8w^3 + 8w^6 = -8+8 (since w^3=1) and we thus have 0.
w^3=1Sorry, how did you get 1+w+w^2=0 just from w^3=1?
You can factorise using . Since w =/= 1 thenSorry, how did you get 1+w+w^2=0 just from w^3=1?
Ahh, thanks, didnt expect to use factorisation :0, guess i need to get used to these question typesw^3=1
w^3-1=0
(w-1)(1+w+w^2)=0
Since w is the non-real root, then w=/=1.
Hence, 1+w+w^2=0.
Using z=x+iy, since Im(z)=2, then z=x+2i.hi guys i need help with this one:
z is any number, such that Im(z) = 2 and z^2 is real. Find z
thanks
thanks kuro that is correct, im kinda confused in "since z^2 is real, 4ix=0" isnt "4ix" the imaginary because of the i ? thank you!Using z=x+iy, since Im(z)=2, then z=x+2i.
Squaring, we get . Since z^2 is real, 4ix=0, therefore x=0.
therefore z=2i?
4ix=0 because if is real then there can be no imaginary component.thanks kuro that is correct, im kinda confused in "since z^2 is real, 4ix=0" isnt "4ix" the imaginary because of the i ? thank you!
oh yeah yeah true, ok thanks friend4ix=0 because if is real then there can be no imaginary component.