Question

Which of the following is a third root of the given complex number? Check all that apply.

64(cos(pi/7)+i sin(pi/7))

Answer 1

you can simplify pi/21 + 2pi*k/3

Answer 2

http://www.wolframalpha.com/input/?i=pi%2F21+%2B+2pi*k%2F3

Answer 3

Is the result what it is simplified above?

Answer 4

$$
\Large{ [64(\cos(\pi/7 + 2\pi k )+i \sin(pi/7 + 2\pi k ))]^{\frac13}
\\ =64^\frac13 [\cos(\frac{\pi/7 + 2\pi k }{3} )+i \sin(\frac{\pi/7 + 2\pi k }{3})]
\\ =4 [\cos(\frac{\pi}{3\cdot7}+\frac{2\pi k}{3})+i \sin(\frac{\pi}{3\cdot7}+\frac{2\pi k}{3})]
\\ = 4 \cdot ( \cos ( \frac{\pi}{21} + \frac{2\pi k}{3}) + i\sin(\frac{\pi}{21} + \frac{2\pi k}{3}))

\\ = 4 \cdot ( \cos ( \frac{\pi}{21} + \frac{7\cdot 2\pi k}{7\cdot 3}) + i\sin(\frac{\pi}{21} + \frac{7\cdot2\pi k}{7\cdot3}))

\\ = 4 \cdot ( \cos ( \frac{\pi}{21} + \frac{14\pi k}{21}) + i\sin(\frac{\pi}{21} + \frac{14\pi k}{21}))

\\ = 4 \cdot ( \cos ( \frac{\pi+14\pi k }{21}) + i\sin(\frac{\pi+14\pi k }{21}))

\\ = 4 \cdot ( \cos ( \frac{\pi(1 +14k) }{21}) + i\sin(\frac{\pi(1+14 k) }{21})

\\~\\
\\ k=0,1,2\\
\begin{cases} \\ =

\end{cases}
}
$$

Answer 5

$$
\Large{ [


4 \cdot ( \cos ( \frac{\pi(1 +14k) }{21}) + i\sin(\frac{\pi(1+14 k) }{21})

\\~\\
\\ k=0,1,2\\
\begin{cases}
4 \cdot ( \cos ( \frac{\pi(1 +14\cdot \color{red}0) }{21}) + i\sin(\frac{\pi(1+14 \cdot \color{red}0) }{21}))
\\ 4 \cdot ( \cos ( \frac{\pi(1 +14\cdot \color{red}1) }{21}) + i\sin(\frac{\pi(1+14 \cdot \color{red}1) }{21}))
\\ 4 \cdot ( \cos ( \frac{\pi(1 +14\cdot \color{red}2) }{21}) + i\sin(\frac{\pi(1+14 \cdot \color{red}2) }{21}))
\end{cases}
}
$$

Answer 6

$$
\Large{ [


4 \cdot ( \cos ( \frac{\pi(1 +14k) }{21}) + i\sin(\frac{\pi(1+14 k) }{21})

\\~\\
\\ k=0,1,2\\
\begin{cases}
4 \cdot ( \cos ( \frac{\pi(\color{red}1) }{21}) + i\sin(\frac{\pi(\color{red}1) }{21}))
\\ 4 \cdot ( \cos ( \frac{\pi( \color{red}{15}) }{21}) + i\sin(\frac{\pi(\color{red}{15}) }{21}))
\\ 4 \cdot ( \cos ( \frac{\pi( \color{red}{29}) }{21}) + i\sin(\frac{\pi(\color{red}{29}) }{21}))
\end{cases}
}
$$

Answer 7

you look at the last post by perl, and match the answer choices with that list.
If the choice is on the list, it's a solution (3rd root of the original expression)

Answer 8

@phi

Are the answers then:

A. 4(cos(π/21) + isin(π/21))

and...

D. 4(cos(15π/21) + isin(15π/21))

Answer 9

yes, those two are on perl's list

Answer 10

@phi its just those two?

Answer 11

you have 4 choices. the choice with a 6 out front in stead of 4 is never going to work
the 12 pi/21 choice also won't work.

There are only 3 cube roots, (if you try k= 3 in perl's answer, you will get pi/21 + 2pi
and the + 2pi doesn't change anything)