How the way of finding roots of a complex number works

Question

I would like to describe my doubt with a question

Finding the sixth roots of $$ 16 - 16\sqrt{3}i $$

One can solve this question by obtaining

$$ z^6 = r^6\exp(i\theta) = 32\exp((-\pi/3 + 2\pi k)/6)$$

Then if you substitute consecutive numbers you get the six roots $$ k = -1, \theta = -7\pi/18$$ $$ k = 0, \theta = -\pi/18$$ $$ k = 1, \theta = 5\pi/18$$ $$ k = 2, \theta = 11\pi/18$$ $$ k = 3, \theta = 17\pi/18$$ $$ k = 4, \theta = 23\pi/18$$ $$ k = 5, \theta = -7\pi/18$$ $$ k = 6, \theta = -\pi/18$$ and so on, where you can see a pattern, i.e. six numbers in a row

And there's my first question, why times $\exp(2\pi k)$ with different ks can obtain different solutions? Isn't that times $\exp(2\pi k)$ to a complex number just rotate the complex number anti-clockwise to the origin position? To my view, the 6 in the denominator seems performing magic here.

For my second question, why there's pattern? Is it somehow related to trigonometric property or modular arithmetic?

Answer 1

For the second question, we have $e^{\frac{2ik\pi}{6}}$ (the is the only part that contains $k$ in it)

So when you reach $k=6$ you'll have $e^{\frac{2i(6)\pi}{6}}= e^{2i\pi}$, and since $$e^{i\theta}=\cos \theta +i\sin \theta$$ Then $e^0=e^{2\pi i}$

From here you can also see that $e^{\frac{2i\pi}{6}}= e^{\frac{2i(7)\pi}{6}}$ ....

So the pattern is formed.

Answer 2

I think you understand how the magnitude works, so let's focus on the angle only and consider finding the $n^{\textrm{th}}$ roots of a number $z=e^{i\theta}$ (where $n$ is a positive integer).

What you are looking for is the set of all numbers which "land" on $z$ after you raise them to the power $n$. These will all have magnitude $1$, so we are asking what numbers end up with angle $\theta$ when you multiply their angles by $n$. In this case only the angle $\theta/n$ becomes $\theta$ after multiplying by $n$, so we have the single value $e^{\theta/n}$.

However, as you know, the choice of the angle $\theta$ to represent $z$ is not unique. All of the values $\theta +2k\pi$ can be used (for integral $k$) as well, since $e^{i\theta}=e^{i\theta + 2k\pi i}$ (simply because the exponential function is periodic with period $2\pi i$).

So we are not only looking for angles that become $\theta$ after they are multiplied by $n$, but also angles that become $\theta + 2k\pi$ (for any integer $k$) after they are multiplied by $n$. As with the original angle, this means the angles in question are $\theta/n + 2k\pi/n$.

There are exactly $n$ of these angles in any (half-open) interval of length $2\pi$. In particular, $$\underbrace{\tfrac{1(2\pi)}n < \tfrac{2(2\pi)}n< \tfrac{3(2\pi)}n <\cdots < \tfrac{n(2\pi)}n = 2\pi}_{n\textrm{ distinct angles in } (0,2\pi]}$$ If you go one more step, you get $2\pi+\tfrac{1(2\pi)}n$, which is exactly $2\pi$ more than the first angle in this sequence, so the points on the circle represented by these angles will repeat again.

Adding $\theta/n$ to these values gets you exactly the values you are after, and you can see that since after exactly $n$ values the angle has increased by $2\pi$, the numbers they represent form a repeating pattern of length $n$.

Thus the $n^{\textrm{th}}$ roots of $e^{i\theta}$ are exactly the numbers $e^{i\theta/n + 2k\pi i/n}$ for any $n$ consecutive values of $k$.