# Plotting roots of unity

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?

• It only makes them easier to place in the Argand-Cauchy plane. – Bernard Jan 21 '20 at 22:50
• I don't understand why the plots would be different in either plane. – user732558 Jan 21 '20 at 23:02
• They're no different. Argand-Cauchy is only the name of the plane when used for the usual representation of complex numbers. – Bernard Jan 21 '20 at 23:05

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 $$cos(6\pi/4)+ isin(6\pi/4)= -i$$.

• Thanks for your answer. I think you meant to say "equilateral triangle" rather than "equilateral circle." Also, I am aware how to plot them; however, I'm not sure why the first question I listed is any different from the second question. – user732558 Jan 21 '20 at 23:42

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.