Further Maths/Lessons/Complex Numbers/Roots of Complex Numbers & Roots of Unity
Core Pure · Complex Numbers · Year 2

Roots of Complex Numbers & Roots of Unity

Every non-zero complex number has exactly $n$ distinct $n$th roots, evenly spaced around a circle — and the special case of the roots of $1$ has a beautiful symmetry all of its own.

16 min read 4 worked examples

Finding the nth roots of a complex number

To solve $w^n=z$ for a given complex number $z$, write $z$ in modulus–argument form. Because $\cos$ and $\sin$ are periodic with period $2\pi$, $z$ can equally be written using any of its infinitely many equivalent arguments $\theta+2k\pi$ for integer $k$ — and it is this freedom that generates every distinct root.

Theorem 6.1 — the n distinct nth roots

If $z = r(\cos\theta+i\sin\theta)$ with $r>0$, then the equation $w^n=z$ has exactly $n$ distinct solutions, given by $$w_k = r^{\frac1n}\left(\cos\frac{\theta+2k\pi}{n}+i\sin\frac{\theta+2k\pi}{n}\right),$$ $$k=0,1,2,\ldots,n-1.$$

JustificationWhy exactly n distinct roots appear, and no more
1
Write $z=r(\cos(\theta+2k\pi)+i\sin(\theta+2k\pi))$ for any integer $k$ (a valid representation of the same $z$, by periodicity).
Every integer k gives an equally valid argument for z.
2
Seek $w=s(\cos\phi+i\sin\phi)$ with $w^n=z$. By De Moivre, $w^n = s^n(\cos n\phi+i\sin n\phi)$.
Write the unknown root in modulus–argument form and apply De Moivre.
3
Matching modulus and argument: $s^n=r \Rightarrow s=r^{1/n}$ (taking the positive real root), and $n\phi=\theta+2k\pi \Rightarrow \phi=\dfrac{\theta+2k\pi}{n}$.
Two real numbers agree iff their modulus and argument match.
4
As $k$ ranges over all integers, $\phi=\dfrac{\theta+2k\pi}{n}$ takes infinitely many values, but they repeat with period $2\pi$ in $\phi$ once $k$ increases by $n$ — so only $k=0,1,\ldots,n-1$ give genuinely distinct points on the Argand diagram.
Angles differing by a multiple of 2π describe the same point.
▪ QED

Geometric interpretation — a regular polygon

Every root $w_k$ has the same modulus $r^{1/n}$, and consecutive roots' arguments differ by exactly $\dfrac{2\pi}{n}$. So all $n$ roots lie on a circle of radius $r^{1/n}$ centred at the origin, spaced at equal angular intervals — the vertices of a regular $n$-gon.

Argand diagram showing the n nth roots of a complex number as equally spaced points forming a regular polygon on a circle of radius r to the power 1 over n
Fig. 1 — the n nth roots of z all lie on a circle of radius r^(1/n), forming the vertices of a regular n-gon, spaced 2π/n apart.
Example 1 · Cube roots of a complex number
Find, in the form $a+bi$, the three cube roots of $8i$.

$8i$ has $r=8$, $\theta=\dfrac{\pi}{2}$. Cube roots: $r^{1/3}=2$, angles $\dfrac{\frac{\pi}{2}+2k\pi}{3}$ for $k=0,1,2$.

$k=0$: $\phi=\dfrac{\pi}{6}$, $w_0=2\left(\cos\dfrac{\pi}{6}+i\sin\dfrac{\pi}{6}\right)=\sqrt3+i$

$k=1$: $\phi=\dfrac{5\pi}{6}$, $w_1=2\left(\cos\dfrac{5\pi}{6}+i\sin\dfrac{5\pi}{6}\right)=-\sqrt3+i$

$k=2$: $\phi=\dfrac{3\pi}{2}$, $w_2=2\left(\cos\dfrac{3\pi}{2}+i\sin\dfrac{3\pi}{2}\right)=-2i$

The three roots $\sqrt3+i,\ -\sqrt3+i,\ -2i$ all have modulus $2$ and are spaced $120°$ apart — the vertices of an equilateral triangle.

Example 2 · Fourth roots forming a square
Find, in the form $a+bi$, the four fourth roots of $-16$.

$-16$ has $r=16$, $\theta=\pi$. Fourth roots: $r^{1/4}=2$, angles $\dfrac{\pi+2k\pi}{4}$ for $k=0,1,2,3$: $\dfrac{\pi}{4},\dfrac{3\pi}{4},\dfrac{5\pi}{4},\dfrac{7\pi}{4}$.

These give $w_k = \sqrt2+\sqrt2i,\ -\sqrt2+\sqrt2i,\ -\sqrt2-\sqrt2i,\ \sqrt2-\sqrt2i$ — four points of modulus $2$, spaced $90°$ apart, forming a square centred at the origin.

Roots of unity

Definition 6.2 — nth roots of unity

The equation $w^n=1$ has $n$ solutions, called the $n$th roots of unity. Since $1=\cos0+i\sin0$ (modulus $1$, argument $0$), Theorem 6.1 gives $$w_k=\cos\frac{2k\pi}{n}+i\sin\frac{2k\pi}{n},$$ $$k=0,1,\ldots,n-1.$$ They lie evenly spaced on the unit circle, one vertex always at $1$ itself ($k=0$), forming a regular $n$-gon.

Writing $\omega=\cos\dfrac{2\pi}{n}+i\sin\dfrac{2\pi}{n}$ (the root with $k=1$, called a primitive $n$th root of unity), De Moivre's theorem shows every other root is a power of $\omega$: $w_k=\omega^k$. So the full set of roots is simply $\{1,\omega,\omega^2,\ldots,\omega^{n-1}\}$.

TheoremThe sum of the nth roots of unity is 0, for n ≥ 2
1
The roots are $1,\omega,\omega^2,\ldots,\omega^{n-1}$, a geometric sequence with first term $1$ and common ratio $\omega$.
Established above using De Moivre.
2
Sum of a geometric series: $1+\omega+\omega^2+\cdots+\omega^{n-1} = \dfrac{\omega^n-1}{\omega-1}$, valid since $\omega\neq1$ for $n\geq2$.
Standard geometric series formula, using ω ≠ 1 to avoid dividing by zero.
3
$\omega^n = \left(\cos\dfrac{2\pi}{n}+i\sin\dfrac{2\pi}{n}\right)^n = \cos2\pi+i\sin2\pi = 1$, by De Moivre.
ω is exactly a primitive nth root of unity, so raising it to the nth power returns to 1.
4
So the sum is $\dfrac{1-1}{\omega-1}=\dfrac{0}{\omega-1}=0$.
Numerator vanishes, and the denominator is non-zero.
▪ QED
Common mistake

Roots of unity questions often ask for the sum excluding $1$ itself — since the total sum of all $n$ roots is $0$, the sum of the remaining $n-1$ non-trivial roots is simply $0-1=-1$, not $0$. Read exactly which roots the question wants summed.

Exam-style questions

Question 1 6 marks

Find, in the form $a+bi$, the three cube roots of $-1$. Show these roots on an Argand diagram, and state the geometric shape they form.

$-1$ has $r=1$, $\theta=\pi$. Cube root angles: $\dfrac{\pi+2k\pi}{3}$ for $k=0,1,2$, i.e. $\dfrac{\pi}{3},\pi,\dfrac{5\pi}{3}$ (M1, A1)

$w_0=\cos\dfrac{\pi}{3}+i\sin\dfrac{\pi}{3}=\dfrac12+\dfrac{\sqrt3}{2}i$; $w_1=\cos\pi+i\sin\pi=-1$; $w_2=\cos\dfrac{5\pi}{3}+i\sin\dfrac{5\pi}{3}=\dfrac12-\dfrac{\sqrt3}{2}i$ (A2)

All three lie on the unit circle, spaced $120°$ apart — the vertices of an equilateral triangle, with one vertex at $-1$ on the negative real axis (B1, B1).

Question 2 5 marks

Prove that the sum of the $n$th roots of unity is $0$ for $n\geq2$. Hence write down the value of $\omega+\omega^2+\omega^3+\omega^4$, where $\omega=\cos\dfrac{2\pi}{5}+i\sin\dfrac{2\pi}{5}$.

Proof as above: the roots $1,\omega,\ldots,\omega^{n-1}$ form a geometric series with ratio $\omega\neq1$, sum $\dfrac{\omega^n-1}{\omega-1}$; since $\omega^n=1$ by De Moivre, the numerator is $0$, giving sum $=0$ (M1, M1, A1).

Here $\omega$ is a primitive $5$th root of unity, so $1+\omega+\omega^2+\omega^3+\omega^4=0$. Since the question asks for the sum excluding the root $1$: $\omega+\omega^2+\omega^3+\omega^4 = 0-1 = -1$ (M1, A1).

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De Moivre's Theorem