Find the cube roots of 27(cos 279° + i sin 279°)
desperately need this! pllleeeaaseee

Question

Find the cube roots of 27(cos 279° + i sin 279°)
desperately need this! pllleeeaaseee

anonymous · Answer

nevermind i figured it out!

zehanz · Answer

Suppose we call your number z, then |z|=27 and arg(z)=279°.
We are looking for numbers, let's call them w, so that w³=z.

Before doing any official calculation, remember: 
taking the third root of a complex number means divide the argument by 3 and take the third root of the modulus (or magnitude, or absolute value, whatever you call it).

Does this help?

anonymous · Answer

i actually got it but thank you! i appreciate it!!

zehanz · Answer

Good for you!

I was all busy writing down my explanation below: (never mind that as well)

Doing it officially now:
Write down two equalities now: 

arg(w³)=3arg(w)=arg(z)=279°, so arg(w)=279°/3=93°
|w³|=|w|³=|z|=27, so |w|=3.

We have now: w=3(cos 93°+ i sin 93°)

Because we have third roots, there are two more solutions: one with an argument of 93°+120°=213° and one with agument 93°+240°=333°

anonymous · Answer

yep got it right!

zehanz · Answer

Well done!