how to use the nth root theorem to find the three roots of z^3 = -343

Question

solomonzelman · Answer

$$(-7)^3=-343$$

anonymous · Answer

there are 3 roots

solomonzelman · Answer

WHAT?!?!  No.... I just gave you a hint.

anonymous · Answer

factor and use the quadratic formula for the two that are not obvious

anonymous · Answer

No I need use the nth theorem to find them.

anonymous · Answer

what on earth is an "nth root theorem"?

anonymous · Answer

demoivre's theorem

anonymous · Answer

k
divide the unit circle in to three equal parts, with one of the parts at $-1$ 
find the complex numbers
multiply the result by $7$

anonymous · Answer

|dw:1401671238378:dw|

anonymous · Answer

Okay I get that but
how do I turn -343 into an angle?

anonymous · Answer

you do not
it is a number, not an angle

anonymous · Answer

finding the cube roots of $-343$ is identical to finding the cube roots of $-1$ except you multiply the results by $7$

anonymous · Answer

if you want the angle for $-1$  look at the unit circle in the complex plane.  it should be pretty clear that the angle is question is $\pi$ or if you are working in degrees, $180$

anonymous · Answer

I don't really understand.
In my book they show how to do it in polar form but I don't know how to get -343 into polar  . Do you know how to convert it because I think that would help.

anonymous · Answer

$$-343=-343+0i$$
The real and imaginary parts, in polar form:
$$\begin{cases}-343=r\cos\theta\
0=r\sin\theta\end{cases}~~\Rightarrow~~r=343,~\theta=\pi$$
since -343 lies on the real axis. $\theta=\pi$ because $r$ can't be negative. So,
$$\large z^3=343e^{i\pi}~~\iff~~z=343^{1/3}e^{i\pi/3}$$
Of course, you also have to account for the complex roots, which have the form
$${\huge 343^{1/3}e^{i\left(\frac{\pi}{3}+\frac{2\pi}{3}\right)}}~~\text{and}~~{\huge 343^{1/3}e^{i\left(\frac{\pi}{3}+\frac{4\pi}{3}\right)}}$$