Complex Roots of Unity

We can use de Moivre's theorem to find solutions to problems like $z^n=1$zn=1. This category of problem is called a complex root of unity problem. 

Let's have a look at how we can use de Moivre's theorem in this way. 

de Moivre's theorem states that if $z=\left(r,\theta\right)$z=(r,θ) then $z^n=\left(r^n,n\theta\right)$zn=(rn,nθ)

So if we want to solve $z^3=1$z3=1 then we start by stating de Moivre's.

Let, $z=\left(r,\theta\right)$z=(r,θ)

then $z^3=\left(r,\theta\right)^3=\left(r^3,3\theta\right)$z3=(r,θ)3=(r3,3θ)

We can write the number $1$1 in modulus argument form like this: $\left(1,2n\pi\right)$(1,2nπ)

So we can write $\left(r^3,3\theta\right)=\left(1,2n\pi\right)$(r3,3θ)=(1,2nπ)

Therefore by equating like parts we can see that:

$r^3=1$r3=1 and $3\theta=2n\pi$3θ=2nπ

$r=1$r=1 and $\theta=\frac{2n\pi}{3}$θ=2nπ3​

 

Now we can find the solutions by letting $n=0,1,2,...$n=0,1,2,...

If $n=0$n=0	$z_1=\left(1,0\right)=1$z1​=(1,0)=1
If $n=1$n=1	$z_2=\left(1,\frac{2\pi}{3}\right)=\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}=-\frac{1}{2}+\frac{\sqrt{3}}{2}i$z2​=(1,2π3​)=cos2π3​+isin2π3​=−12​+√32​i
If $n=2$n=2	$z_3=\left(1,\frac{4\pi}{3}\right)=\cos\frac{4\pi}{3}+i\sin\frac{4\pi}{3}=-\frac{1}{2}-\frac{\sqrt{3}}{2}i$z3​=(1,4π3​)=cos4π3​+isin4π3​=−12​−√32​i

So the $3$3 solutions to $z^3=1$z3=1 are$1,-\frac{1}{2}+\frac{\sqrt{3}}{2}i,-\frac{1}{2}-\frac{\sqrt{3}}{2}i$1,−12​+√32​i,−12​−√32​i

 

As you will have discovered through the activity above, roots of unity occur in conjugate pairs and are evenly spaced around the plane.  For $z^n=1$zn=1, they are evenly spaced by an angle of $\frac{2\pi}{n}$2πn​.

Roots of unity are cyclic around the region, starting position equal to $\frac{2\pi}{n}$2πn​ if $z^n=1$zn=1 and $\frac{\pi}{n}$πn​ if $z^n=-1$zn=−1.

 

Worked Examples

Question 1

Question 2

Question 3