For $z^2=1$ there is a formula. But for powers higher than two I do not know what to do.
I was trying to replace $z$ by $a+ib$ and calculate the cube. This probably works. But then problem is that the next exercise I have to do is to solve $z^6=1$. There must be a smarter way.
Kind regards,
On
A smarter way to solve $z^6=1$ once you have solved $z^3=1$:
if $z^6=1,$ then $z^3=1$ or $(-z)^3=1$
On
Really the way to do this is using the complex exponential. $1 = e^{2n\pi i}$ for $n \in \mathbb{Z}$. Suppose $z$ is a solution to $z^m = 1$. You can convince yourself that $|z| = 1$. Then $z =e^{ix}$ for some $x$, so $z^m = e^{imx} = 1 = e^{2\pi i}$. Elementary arguments then give you that the set of solutions is then $e^{2\pi i n/m}$ for $n \in \mathbb{Z}$.
On
The solutions are the three third roots of unity: $\{1,\omega,\omega^2\}$, where $\omega=e^{2\pi i/3}$.
On
The equation $z^n = 1$ for cases $$n = 2,3,5$$ and a couple more can be solved by factoring using the (generalized) conjugare rule and solving a quadratic equation without resorting to trigonometry or other representations. So let's explore the case n = 3
The generalized conjugare rule is, as it's name implies a generalization of the conjugate rule
$$a^2 - b^2 = (a - b)(a + b)$$
But where you have other exponents than 2. In the case of cubes and you can factor out (a - b) in much the same way
$$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$$
and most generally
$$a^n - b^n = (a - b)(a^{n - 1} + a^{n-2}b + a^{n-3}b^2 + ... + a b^{n - 2} + b^{n - 1}$$
Let's solve $$z^3 = 1$$ by using the cubic pattern
$$z^3 - 1 = 0$$
$$z^3 - 1^3 = 0$$
$$z^3 - 1^3 = (z - 1)(z^2 + z + 1) = 0$$
Looking at this factorization we find the first root $z_1 = 1$ immediately and the two remaining roots can be found by solving
$$z^2 + z + 1 = 0$$
but that's just a quadratic eequation and you can solve that one using a plethera of different methods.
The solutions are evenly spaced around a circle of radius one, so you can use sin and cos for this. The solutions to $z^n = 1$ are
$$ z = \cos\left(\frac{i}{n} 2 \pi\right) + i \sin\left(\frac{i}{n} 2 \pi\right)$$ where $i$ takes on the value $0$ up to $n-1$.