Question

CUBE ROOTS OF -8: I find that r =sqrt(-8^2 +0)= 8. And Arg IzI = pi or 180 degrees.
Now, when k=0, I am getting 2[ cos0 + iSin0), when writing in complex form this will give me 2, cos 0=1! or am I missing something, because, 2 x 2 x 2 =8 not -8?

Answer 1

\(\large - 8 = 8(\cos \pi + i \sin \pi) = 8e^{i \pi }\)

Answer 2

cube roots of -8 = \(\large \sqrt[3]{8} ~e^{i \frac{\pi + 2k \pi}{3} } = 2 ~e^{i \frac{\pi + 2k \pi}{3} } = 2 \left[ \cos ( \frac{\pi + 2k\pi}{3}) + i \sin ( \frac{\pi + 2k\pi}{3}) \right] \)

\(0 \le k \lt 3 \)

Answer 3

so when k =0?

Answer 4

that is the part that i don't understand

Answer 5

the result i get in complex form is 2(1 +0i)

Answer 6

@ganeshie8

Answer 7

\[\large 2 \left[ \cos ( \frac{\pi + 2k\pi}{3}) + i \sin ( \frac{\pi + 2k\pi}{3}) \right]\]

plugin \(k = 0\)

\[\large 2 \left[ \cos ( \frac{\pi }{3}) + i \sin ( \frac{\pi}{3}) \right]\]

Answer 8

thats another cube root for -8,
cube it and see if u get -8

Answer 9

http://www.wolframalpha.com/input/?i=%282%28cos%28pi%2F3%29+%2B+isin%28pi%2F3%29%29%29%5E3