the cube roots of 27(cos 330° + i sin 330°) will be </span>∛[27(cos 330° + i sin 330°)]
we know that e<span>^(ix)=cos x + isinx therefore </span>∛[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 <span>therefore </span> 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°] <span> the answer is
