What are the roots of $z^n=-1$?

Question

It is obvious that the roots of $z^n=1$ are $\eta_{k}=e^\frac{i2\pi k}{n},\,\, $where $k=0,1,2...n-1.$ I am wondering what are the roots of $z^n=-1.$ I first thought these should be $\eta_{j}e^{i\pi}, \,\,$since $e^{i\pi}=-1.$ But it proved to be false. I will appreciate a suggestion.

Thanks.

Answer 1

Hint: Since $z^{2n}-1=(z^n-1)(z^n+1)$, the roots of $z^n=-1$ are the roots of $z^{2n}=1$ that are not a root of $z^n=1$.

Answer 2

You can take your favorite $n-$root of $-1$, for instance, you may take

$$\rho=e^\frac{i\pi}{n}$$

Notice that $\rho^n=e^\frac{in\pi}{n}=e^{i\pi}=-1$.

Now: $\big(\eta_j\rho\big)^n=\eta_j^n\rho^n=-1$ for any $j\in\{0,\cdots,n-1\}$

Answer 3

$$z^n=-1=e^{i(2k+1)\pi}$$ $$z=e^{i\frac{2k+1}{n}\pi}$$

Answer 4

For $z=re^{i\theta}$, $z^n=r^ne^{in\theta}$.

If $z^n=1$ then $z=(-1)^{1/n}=(e^{i\pi})^{1/n}=e^{\frac{i\pi}{n}}$.

Now, of course we expect $n$ roots, and those come from the fact that $e^{i\theta}=e^{i(\theta+2k\pi)}, \quad k\in\mathbb{N}$ (intuitively, if you "rotate" the number by $360 ^\circ$).

Hence all the roots are of the form $e^{i\frac{2k+1}{n}\pi}, \quad k \in \{0,\dots,n-1\}$.