Need here asap 6. find the cube roots of 27(cos 330° + i sin 330°).

we have that

the cube roots of 27(cos 330° + i sin 330°) will be
∛[27(cos 330° + i sin 330°)]

we know that
e^(ix)=cos x + isinx
therefore
∛[27(cos 330° + i sin 330°)]> ∛[27(e^(i330°))]> 3∛[(e^(i110°)³)]
3∛[(e^(i110°)³)]> 3e^(i110°)> 3[cos 110° + i sin 110°]

z1=3[cos 110° + i sin 110°]

 cube root in complex number, divide angle by 3
360nº/3 = 120nº --> add 120º for z2 angle, again for z3
therefore

z2=3[cos ((110°+120°) + i sin (110°+120°)] > 3[cos 230° + i sin 230°]

z3=3[cos (230°+120°) + i sin (230°+120°)]> 3[cos 350° + i sin 350°]

the answer is 

z1=3[cos 110° + i sin 110°]

z2=3[cos 230° + i sin 230°]

z3=3[cos 350° + i sin 350°]

Your answer is gonna be zero

Step-by-step explanation: