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Firlakuza [10]
3 years ago
14

NEED HELP HERE ASAP 6. Find the cube roots of 27(cos 330° + i sin 330°).

Mathematics
1 answer:
VLD [36.1K]3 years ago
7 0
<span>we have that

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 

</span>z1=3[cos 110° + i sin 110°]<span>

</span>z2=3[cos 230° + i sin 230°]

z3=3[cos 350° + i sin 350°]<span>

</span>
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