Plotting roots of unity

Question

In an assignment, I have the following two exercises:

1. Draw a Cartesian Coordinate plane with the unit circle centered at the origin. Show where each of the three third roots of unity lie on the unit circle

2. Draw a Cartesian coordinate plane with the unit circle centered at the origin. Show where each of the three third roots of unity lie on the unit circle, where each root is written in the form $e^{ix}$ (where $e^{ix} = \cos(x) + i\sin(x)$).

I don't understand why the solution to these two problems would be any different. I know what the roots are, but how would rewriting them from $a + bi$ form to exponential form change where they lie?

Answer 1

The nth roots of unity lie on the circle with center at the origin with radius 1 and are the vertices of a regular polynomial with n vertices. Of course 1 is one of those vertices so you can mark the others by measuring off angle of size $\frac{2\pi}{n}$.

So, for example, the third roots of unity are the vertices of an equilateral circle, lying at 1, $cos(2\pi/3)+ isin(2\pi/3)= -0.5+ i\sqrt{3}/2)$, and $cos(4\pi/3)+ isin(4\pi/3)= -0.5- i\sqrt{3}/2)$.

The fourth roots of unity are the vertices of a square, lying at 1, $cos(2\pi/4)+ isin(2\pi/4)= i$, $cos(4\pi/4)+ isin(4\pi/4)= -1$, and $cos(6\pi/4)+ isin(6\pi/4)= -i$.

Answer 2

If we take $z=1$ as $z = 1 + 0i$, $|z| = \sqrt {1^2 + 0^2} = 1$, $\arg z = \arctan \frac 01 = \pi$ and using DeMoivre's Theorem for the roots of unity $$1^{1/n} = \left(\cos \frac {k\theta}{n} + i \sin \frac {k\theta}{n}\right), k \in (0, 1, 2 \dots n)$$ we have $$1^{1/3} = \left(\cos \frac {k \pi}{3} + i \sin \frac {k \pi}{3}\right), k \in (0,1,2)$$

In the first case, you're plugging in the values of $k$ and finding the sine and cosine of each multiple angle to get $$1, -\frac {1}{2} + \frac {i \sqrt{3}}{2}, -\frac {1}{2} - \frac {i \sqrt{3}}{2}$$

In the second case, you're rewriting $\cos \frac {k \pi}{3} + i \sin \frac {k \pi}{3}$ as $e^{i k \pi/3}$ and letting $k = 0, 1, 2$; so the third roots of unity in exponentials are $$e^{i(0\pi/3)} (= 1), e^{i(\pi/3)}, e^{i(2\pi/3)}$$

Hence, the cartesian and the exponential forms are equivalent, but it's quicker to write the exponential form.