Using de Moivre to solve $z^3=-1$, one solution is $\cos \pi+i \sin \pi$. What am I doing wrong?

Question

I have a problem in my calculations using de Moivre for complex numbers: I have been staring at this for an hour, without success. Can anybody please tell me what I am doing wrong?

I have to calculate $z^3=-1$

So, besides the obvious solution $z=-1$:

$$z^3=r^3(\cos3\theta+i \sin3\theta)=1(\cos\pi + i \sin \pi)$$

so $r = 1$ and $3\theta=\pi+2k\pi$, so $$θ = \frac{\pi}{3}+\frac{2}{3}k\pi$$

The first solution that comes out of this is correct, namely $\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}$. However, the second solution that comes out is patently wrong: $\cos \pi+i \sin \pi$. The third one would be right again: $\cos \frac{5\pi}{3}+i \sin \frac{5\pi}{3}$, which would be our final one. I do not understand why the second one came out wrong and what I should change in the procedure so that I do not make this mistake again.

The error is obviously located somewhere in the $θ = \frac{π}{3}+\frac{2}{3}kπ$, but this is how the textbook instructs me to go through the procedure.

Any thoughts would be really appreciated...

Answer 1

Since $\cos(\pi)+i\sin(\pi)=-1$ and since you wrote that $-1$ is a solution, I fail to see what's the problem.

Answer 2

The second solution ($\cos \pi + i \sin \pi$) is correct (part of the root for $z^3 =-1$) as $\cos \pi$ is $-1$ and $\sin \pi$ is $0$, hence you have $-1+0i$ = $-1.$ From a unit circle perspective, $\cos \theta$ is decreasing and negative and $\sin \theta$ is decreasing but positive in the second quadrant.