given that z=1-root3i solve equation z^6+4(1-i)z^3+p+qi=0

Question

complex numbers

Answer 1

z⁶=(1-i√3)⁶=1-6i√3+15(-i√3)²+20(-i√3)³+15(-i√3)⁴+6(-i√3)⁵+(-i√3)⁶,

(-i)²=-1; (-i)³=i; (-i)⁴=1; (-i)⁵=-i; (-i)⁶=-1;

z⁶=1-6i√3-45+60i√3+135-54i√3-27=64;

z³=1-3i√3+3(-i√3)²+(-i√3)³=1-3i√3-9+3i√3=-8.

[By de Moivre:

z=ae^niθ=a(cos(nθ)+isin(nθ)).

z=a(cosθ+isinθ)=1-i√3, acosθ=1, asinθ=-√3, a²=1+3=4, a=2; tanθ=-√3, θ=-π/3.

z=2(cos(-π/3)+isin(-π/3))=2(½-½√3)=1-√3✔️ (n=1)

z³=2³(cos(-π)+isin(-π))=8(-1+0)=-8; z⁶=64 (n=3)]

z⁶+4(i-1)z³+p+qi=0 becomes 64+4(i-1)(-8)+p+qi=0,

64-32i+32+p+qi=0, 96-32i+p+qi=0.

p=-96, q=32.