Find the cube roots of 27(cos 330 + i sin 330)

answer:

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°]

Step-by-step explanation:

x = -3

y = -18

step-by-step explanation:

to solve by substitution, substitute one variable and solve for the other. then substitute again to find the remaining variable.

x-1/3(3x-9) = -3 since y=3x-9 substittute y.

x-1/3(3x-9)=-3

x + x +3 = -3

2x+3=-3

2x+3-3= -3-3

2x=-6

x=-3

substitute again into y=3x-9

y=3(-3)-9

y=-9-9

y=-18