Nth roots of complex numbers: formula, completeness, and examples

Content updated: October 22, 2026

An nth root of a complex number is any number whose nth power equals that number. For a nonzero number, the roots have the same modulus and equally spaced arguments; for zero, there is only one root value.

Learning goals: derive the root formula from de Moivre’s formula, prove that the list is complete and its roots are distinct, and check roots by raising them to the required power.

Prerequisites: modulus, arguments modulo 2π, and de Moivre’s formula.

On this page: Definition · Method · Examples · Practice · Further learning

1. The root problem and its assumptions

Let $n\ge1$ be an integer. Finding the $n$th roots of $z$ means solving $w^n=z$ in $\mathbb C$. If $z=0$, the only solution is $w=0$, since $|w|^n=0$ forces the modulus to be zero. In this case, there are not $n$ distinct root values.

Suppose that $z\ne0$ and write $z=r(\cos\varphi+i\sin\varphi)$, where $r>0$. Every solution $w$ is nonzero, so we may write $w=s(\cos\alpha+i\sin\alpha)$ with $s>0$. By de Moivre’s formula, the power equation is equivalent to $s^n=r$ and $n\alpha=\varphi+2k\pi$ for some integer $k$.

2. The formula, distinctness, and completeness

Each number in the list has modulus $\sqrt[n]r$, and its $n$th power has argument $\varphi+2k\pi$, so every listed number is a solution. If two indices $k,\ell$ determine the same root, the difference of their arguments, $2(k-\ell)\pi/n$, is an integer multiple of $2\pi$. Thus $n$ divides $k-\ell$. Since $|k-\ell|<n$, it follows that $k=\ell$. Hence the list contains $n$ distinct roots.

Conversely, every solution has an argument determined by an integer $k$ as above. Taking the remainder of that index modulo $n$ reduces it to exactly one of $0,\ldots,n-1$, so the list is complete. Replacing the argument $\varphi$ by $\varphi+2m\pi$ merely reindexes the roots; it does not change their set.

Geometrically, the roots lie on the circle with center $O$ and radius $\sqrt[n]r$, with angular spacing $2\pi/n$. For $n\ge3$, they are the vertices of a regular polygon; for $n=2$, they are two points symmetric about the origin; for $n=1$, the sole root is $z$ itself. These facts follow from the formula, rather than being assumptions inferred from the diagram.

3. Three worked examples

Example 1. The square roots of 1 + √3 i

The number $z=1+\sqrt3i$ has modulus 2 and an argument $\pi/3$. Its two square roots have modulus $\sqrt2$ and arguments $\pi/6,7\pi/6$. Therefore

Squaring the first root gives real part $6/4-2/4=1$ and imaginary part $2(\sqrt6/2)(\sqrt2/2)=\sqrt3$. The second root is its negative and therefore has the same square. Their points in the complex plane are symmetric about the origin.

Example 2. The fourth roots of −8 + 8√3 i

We have $z=16[\cos(2\pi/3)+i\sin(2\pi/3)]$. The roots have modulus 2 and arguments $\pi/6+k\pi/2$ for $k=0,1,2,3$. In that order, the four roots are

The fourth power of the first root is $-8+8\sqrt3i$. Each successive root is obtained by multiplying by $i$, which leaves the fourth power unchanged because $i^4=1$. Their arguments differ by $\pi/2$, so the roots form the four vertices of a square on the circle of radius 2. Numbering them from 1 to 4, as in the earlier presentation, only changes the order of the list.

Example 3. The cube roots of 1

For $z=1$, the modulus is 1 and an argument is 0, not $\pi/2$. The three root arguments are 0 and $2\pi/3,4\pi/3$, so

A further check follows from the factorization $w^3-1=(w-1)(w^2+w+1)$. All three roots have cube equal to 1. They are the vertices of an equilateral triangle on the unit circle. Choosing the argument $\pi/2$ for the original number would instead give the root problem for $i$, which is a different problem.

4. Practice

1. Find all fourth roots of 1.
2. How many distinct cube root values does 0 have?
3. Find the cube roots of $-1$.

5. Common mistakes

Keep the assumption that $n$ is a positive integer. Do not assign an argument to zero. Do not use infinitely many integer indices $k$ and count repeated values as new roots. The set of roots does not depend on the chosen representative argument, but the modulus and an argument of the original number must be determined correctly.

6. Further learning

Basic properties of modulus and argument (English lesson) · Representations of complex numbers (English lesson) · Complex numbers (English lesson) · Roots of a quadratic equation (English lesson).