# How does one simplify exponents for complex primitive nth roots of unity?

Let us define a complex primitive N-th root of unity, omega: $$\omega = \cos(\theta) + i\sin(\theta) \\ = e^{\frac{2\pi}{N}}$$

By the definition of an nth root of unity, ω is the second solution to the equation x^N = 1. Can someone explain to me how I can simplify the solutions' exponents via modular arithmetic? I sort of understand the generic congruence modulo nonsense (for M > N, Q is some quotient and R is some remainder):

$$\omega^M; \\ M = QN + R \\ \omega^{QN + R} \\ = \omega^{QN} + \omega^R \\ = ω^R \\ R \equiv M \mod N$$

I don't exactly understand this. This is how my quantum mechanics professor explained it. I need to input 6 values corresponding to ω (those values are irrelevant to the problem), however I can only use ω and ω^2, even though N = 6. How do I get rid of the other exponents if my problem includes ω^3, ω^6, ω^9, ω^12 and ω^15?

Your calculations should read $$w^{QN+R} = w^{QN}\times w^R = (w^N)^Q\times w^R = (1)^Q\times w^R = w^R$$ The idea is simple - each root of unity represents a point on the unit circle $\{z\in \mathbb{Z} : |z| = 1\}$. For a fixed $N \in \mathbb{N}$, there are $N$ equidistant points, (including 1), each one at an angle of $2\pi k/N$ radians from the $X$-axis (for $1\leq k\leq N$). When you multiply two such points, you merely add these angles.

In the case $N=6$, $$w = e^{2\pi i/6} = e^{\pi i/3}$$ corresponds to the point at 60 degrees from the $X$-axis.

To compute $w^3$, you merely add $60^{\circ}$ 3 times to get $180^{\circ}\leftrightarrow -1$ Similarly, $$w^6 \leftrightarrow 360^{\circ} \Rightarrow w^6 = 1$$ $$w^9 \leftrightarrow (60*9) = 540^{\circ} \equiv 180^{\circ} \Rightarrow w^9 = -1$$ and so on.

Hope this helps

Well, for this you should realize that whenever the value $M$ is greater than $N$, you're coming back to a point which you already visited earlier. But what point is it exactly is what you need to find out. For example, if $N=12$, then if you're asked to find out $\omega^{15}$, then $\omega^{15}=\omega^{12+3}$. Now we already know that $\omega^{12}=1$and the angle is $0$; hence what you're required to find out is $\omega^{3}$. Now the math notation that you've written follows this example. This is a simple consequence of Phasor rotation on a circle.

Given $M$, $M$ can be written as a multiple of $N$ and a remainder. The remainder is what is necessary because you are just moving around the circle in a complex plane whose magnitudes are all 1 but their phases are different. The phase of the required $M$ is what is given by $R$.

• Great. The relationship of 15 = 12 + 3 makes a lot simpler, in addition to Prahlad's answer. – cygorx Sep 23 '13 at 2:22