Why does the De Moivre formula work?

Question

In order to find the nth roots of a complex number you need to use this formula.

https://en.wikipedia.org/wiki/De_Moivre%27s_formula#Roots_of_complex_numbers

For me I'm trying to understand where this formula comes from. I get that they try to write the complex number in polar form so that it makes it easy to take the power, but I'm not sure where the $2\pi k$ where $k = 0,...,n-1$ part comes from.

I was wondering if someone could intuitively explain what the $2\pi k$ part does and why we need it.

Answer 1

$2\pi$ is a complete circle.
Dividing it into $n$ parts gives $\dfrac {2\pi}n$.
$\dfrac{2\pi k}n$, where $k=0,1,2,\cdots, n-1$, gives the positions of the $1$st, $2$nd, .., $n$th parts respectively.

Answer 2

Hint:

cosider $$x_1 \angle \theta_1 \times x_2 \angle \theta_2 = x_1 x_2 \angle (\theta_1+\theta_2+2 k \pi)$$

Answer 3

It comes from the fact that there are $n$ possible solutions to the equation $$ z^n = 1 $$ when you allow $z$ to be a complex number: $$ 1, e^{2\pi i/n}, e^{4\pi i/n}, \ldots, e^{2\pi k i/n},\ldots, e^{2\pi(n-1) i/n}. $$ Then applying $$ e^{i\theta} = \cos\theta + i\sin\theta $$ to this above list give you the formula in the Wikipedia article, sans the modulus (which can be accounted for easily).

Answer 4

Consider some complex number given in polar form $z=re^{i\phi}$. Now what happens when you evaluate $ze^{2\pi ik}= re^{i(\phi+2\pi k)}$ for some integer $k$?

$$ze^{2\pi ik}=re^{i(\phi+2\pi k)}= re^{i\phi}e^{2\pi ik} = re^{i\phi}\cdot 1= z$$

Whenever we add some integer multiple of $2\pi i$ to the argument of a complex number... we have the same complex number! Well, now let's see what happens when we take a root of that number.

$$\begin{align} z^{1/n}&=r^{1/n}e^{\frac{i(\phi+2\pi k)}{n}}\\ &=r^{1/n}\left\{\cos\left({\frac{\phi+2\pi k}{n}}\right)+i\sin\left({\frac{\phi+2\pi k}{n}}\right)\right\}\\ \end{align}$$

It appears that the $2\pi k$ term doesn't vanish anymore, since it's being divided by $n$. Furthermore, while $z$ could take any $k$ and remain the same, this is no longer the case for the root of $z$! The root now depends very much on the exact argument of the complex number, $2\pi k$ included.

Notably, any integer root $n$ will result in $n$ different points, uniformly distributed about the origin. For example, here's a plot of the fourth root of unity, $1^{1/4}$.

We have four distinct points: $z= \pm 1,\pm i$. These correspond to $k=0,1,2,3$. If you continue to plot points for $k=5,6,7,...$ you'll just repeat these points.