Question

find the cube roots of the complex number 27(cos 11pi/6+i sin 11pi6)

Answer 1

take the cube root of 27, get 3
divide the angles by 3

Answer 2

...

Answer 3

is that 33pi/0?

Answer 4

that is not the answer im looking for....

Answer 5

\[
\frac {33\pi} 6
\]

Answer 6

my answer is suppose to be 3(cos 11pi/18 + i sin 11pi/18)

Answer 7

Sorry, I did
\[
x^3
\]
instead 0f
\[x^{\frac 1 3}
\]

Answer 8

and 3(cos 23pi/18 + i sin 23pi/18) and 3(cos 35pi/18 + i sin 35pi/18)

Answer 9

ok

Answer 10

\[ x = 27 e^{ i \left( \frac {11 \pi}{ 6}+ 2 k \pi
\right)
}\\
x^{\frac 1 3}= 27^{\frac 1 3 }
e^{ i \left( \frac {11 \pi}{ 18}+ 2 k\frac \pi 3
\right)
}\\
\]

Answer 11

Take k=0,1,2 and you get your three answers.

Answer 12

@lorim9, you got this?

Answer 13

i think i do now.

Answer 14

divide \(\frac{11\pi}{6}\) by 3, get \(\frac{11\pi}{18}\)
add \(2\pi\) to \(\frac{11\pi}{6}\) get \(\frac{23\pi}{6}\) divide by 3 again get \(\frac{23\pi}{18}\)

Answer 15

what you and the other person is kinda similar?

Answer 16

and once more, add \(2\pi\) to \(\frac{23\pi}{6}\) get \(\frac{35}{6}\) divide once more and get \(\frac{35}{18}\) those are your three angles.
yes it is exactly the same

Answer 17

alright thank you guys!