In short

The nth roots of a complex number w = R(\cos\phi + i\sin\phi) are the n solutions of z^n = w. They are

z_k = R^{1/n}\left(\cos\frac{\phi + 2k\pi}{n} + i\sin\frac{\phi + 2k\pi}{n}\right), \quad k = 0, 1, \dots, n-1.

These n roots are equally spaced on a circle of radius R^{1/n}, forming a regular n-gon. They differ from each other by multiplication by the roots of unity — once you find one root, the rest are that root times 1, \omega, \omega^2, \dots, \omega^{n-1}.

You know how to find the nth roots of 1 — they sit equally spaced on the unit circle. But what about the nth roots of an arbitrary complex number? What are the cube roots of 8i, or the fourth roots of -4?

The answer turns out to be almost as clean. The roots still form a regular polygon — just on a different circle, rotated to a different starting angle. The roots of unity are the special case; the general case is one small step further.

The formula

Start with a non-zero complex number w in polar form:

w = R(\cos\phi + i\sin\phi)

where R = |w| > 0 and \phi = \arg(w). You want all z such that z^n = w.

Write z in polar form too: z = r(\cos\theta + i\sin\theta). By De Moivre's theorem:

z^n = r^n(\cos(n\theta) + i\sin(n\theta))

Set this equal to w = R(\cos\phi + i\sin\phi) and match modulus and argument:

Modulus: r^n = R, so r = R^{1/n} (the unique positive real nth root of R).

Argument: n\theta = \phi + 2k\pi for integer k, so \theta = \frac{\phi + 2k\pi}{n}.

As k runs from 0 to n-1, you get n distinct values of \theta, each separated by \frac{2\pi}{n}. At k = n, the angle wraps around to \theta_0 + 2\pi, which is the same point as k = 0.

$n$th roots of a complex number

Let w = R(\cos\phi + i\sin\phi) with R > 0. The n solutions of z^n = w are

z_k = R^{1/n}\left(\cos\frac{\phi + 2k\pi}{n} + i\sin\frac{\phi + 2k\pi}{n}\right), \quad k = 0, 1, \dots, n-1.

In exponential form: z_k = R^{1/n}\, e^{i(\phi + 2k\pi)/n}.

Three observations jump out from this formula:

  1. All n roots have the same modulus R^{1/n}. They lie on a circle of radius R^{1/n}, not necessarily the unit circle.

  2. The roots are equally spaced — consecutive roots differ by an angle of 2\pi/n, the same spacing as the nth roots of unity.

  3. They form a regular n-gon centred at the origin, rotated so that the first root (k = 0) sits at angle \phi/n.

The four fourth roots of a complex number w forming a square on a circle of radius R to the one-fourthA complex plane showing a circle of radius R to the one-fourth centred at the origin. Four points are equally spaced on this circle, forming a square. The first root z0 is at angle phi over 4, and subsequent roots are at angles phi over 4 plus pi over 2, phi over 4 plus pi, and phi over 4 plus three pi over 2. Lines from the origin to each root are drawn. The original number w is shown on a larger circle of radius R. ReIm R R^(1/4) φ/4 z₀ z₁ z₂ z₃ w φ The four roots (solid circle) form a square. The original w sits on the dashed circle.
The four fourth roots of $w$ (shown on the solid circle of radius $R^{1/4}$). They form a square, with the first root $z_0$ at angle $\phi/4$. The original number $w$ sits on the dashed circle of radius $R$ at angle $\phi$. Raising any root to the fourth power multiplies the modulus to $R$ and the angle to $\phi$ (or $\phi + 2k\pi$), landing back at $w$.

The connection to roots of unity

Here is the cleanest way to think about it. Let z_0 be any one particular nth root of w — say the one with k = 0:

z_0 = R^{1/n}\, e^{i\phi/n}

Then all the nth roots of w are:

z_0, \; z_0\omega, \; z_0\omega^2, \; \dots, \; z_0\omega^{n-1}

where \omega = e^{2\pi i/n} is a primitive nth root of unity.

Check: (z_0 \omega^k)^n = z_0^n \cdot (\omega^k)^n = w \cdot \omega^{kn} = w \cdot 1 = w. Each one is indeed an nth root of w.

This is the key insight: finding the nth roots of any complex number reduces to finding one root and then multiplying by the nth roots of unity. The roots of unity act like a "rotational offset" that generates all the others from a single starting point.

How one root and the roots of unity generate all nth rootsA diagram showing three stages. On the left, a single root z0. In the middle, the three cube roots of unity 1, omega, omega squared on a small unit circle. On the right, the three cube roots of w formed by multiplying z0 by each root of unity, shown on a circle of radius R to the one-third. one root z₀ R^(1/n) × roots of unity 1 ω ω² unit circle = all n roots z₀ z₀ω z₀ω² R^(1/n) One root times each root of unity gives all n roots.
The structure of $n$th roots. Find one root $z_0$, then multiply by each of the $n$th roots of unity. The result is all $n$ roots of $w$, equally spaced on a circle of radius $R^{1/n}$. For cube roots ($n = 3$), this means one root generates two more by rotation through $120°$ and $240°$.

Geometric interpretation

The n roots of a complex number w always form a regular n-gon:

This means you can see the roots before computing them. Given w, plot it on the complex plane. The roots sit on a smaller circle (radius = nth root of the distance), rotated so that the angle is divided by n, and replicated n times around the circle.

When w is a positive real number — say w = 8 — the angle \phi = 0, so the first root sits on the positive real axis at 8^{1/3} = 2 (for cube roots). The other roots are at 120° and 240°:

z_0 = 2, \quad z_1 = 2\omega = -1 + i\sqrt{3}, \quad z_2 = 2\omega^2 = -1 - i\sqrt{3}

Only one of these roots is real. The other two are complex conjugates. This is a general pattern: a positive real number has exactly one real nth root (when n is odd) or two real nth roots (when n is even, \pm R^{1/n}), and the remaining roots come in conjugate pairs.

Applications

Solving polynomial equations

The equation z^4 = -16 is a fourth-degree polynomial equation. Writing -16 = 16(\cos\pi + i\sin\pi), the four roots are:

z_k = 16^{1/4}\left(\cos\frac{\pi + 2k\pi}{4} + i\sin\frac{\pi + 2k\pi}{4}\right) = 2\left(\cos\frac{(2k+1)\pi}{4} + i\sin\frac{(2k+1)\pi}{4}\right)

For k = 0, 1, 2, 3, the angles are \pi/4, 3\pi/4, 5\pi/4, 7\pi/4 — the four diagonal directions. In Cartesian form:

z_0 = \sqrt{2} + i\sqrt{2}, \quad z_1 = -\sqrt{2} + i\sqrt{2}, \quad z_2 = -\sqrt{2} - i\sqrt{2}, \quad z_3 = \sqrt{2} - i\sqrt{2}

These are the vertices of a square with side length 2\sqrt{2}, rotated 45° from the axes.

Factoring polynomials

The identity z^n - w = (z - z_0)(z - z_1) \cdots (z - z_{n-1}) factors any binomial z^n - w into linear factors over the complex numbers. For z^3 - 8:

z^3 - 8 = (z - 2)(z - 2\omega)(z - 2\omega^2) = (z - 2)(z^2 + 2z + 4)

The quadratic factor z^2 + 2z + 4 is irreducible over the reals — its roots 2\omega and 2\omega^2 are complex conjugates with discriminant 4 - 16 = -12 < 0.

Interactive: exploring nth roots

Drag the red point along the curve below. The readouts show the coordinates (x, y) and the modulus r = \sqrt{x^2 + y^2}. As you move the point, think about where the cube roots of that number would sit — on a circle of radius r^{1/3} at one-third of the argument.

Interactive complex number explorer for nth rootsA complex plane from negative 5 to 5 on both axes. A red draggable point can be moved freely. Readouts show the real part, imaginary part, modulus, argument, and the cube root modulus updating in real time. ReIm 1234 −1−2−3−4 drag the red point to explore
Drag the red point to any position on the complex plane. The readouts show the modulus $|w|$, the argument $\arg(w)$, and the cube-root modulus $|w|^{1/3}$. The cube roots of the number at the red point would sit on a circle of that cube-root radius, at angles $\arg(w)/3$, $\arg(w)/3 + 2\pi/3$, and $\arg(w)/3 + 4\pi/3$.

Worked examples

Example 1: Find the cube roots of $8i$

A complex number with a clean modulus and argument, giving roots with exact trigonometric values.

Step 1. Write 8i in polar form.

|8i| = 8, \qquad \arg(8i) = \frac{\pi}{2}

So 8i = 8\left(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\right).

Why: 8i sits on the positive imaginary axis, at distance 8 from the origin and angle \pi/2 from the real axis.

Step 2. Apply the nth root formula with n = 3, R = 8, \phi = \pi/2.

z_k = 8^{1/3}\left(\cos\frac{\pi/2 + 2k\pi}{3} + i\sin\frac{\pi/2 + 2k\pi}{3}\right) = 2\left(\cos\frac{\pi + 4k\pi}{6} + i\sin\frac{\pi + 4k\pi}{6}\right)

Why: 8^{1/3} = 2. The angle \frac{\pi/2 + 2k\pi}{3} = \frac{\pi + 4k\pi}{6} for each k = 0, 1, 2.

Step 3. Compute each root.

k = 0: \theta_0 = \pi/6.

z_0 = 2\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right) = 2\left(\frac{\sqrt{3}}{2} + i \cdot \frac{1}{2}\right) = \sqrt{3} + i

k = 1: \theta_1 = \pi/6 + 2\pi/3 = 5\pi/6.

z_1 = 2\left(\cos\frac{5\pi}{6} + i\sin\frac{5\pi}{6}\right) = 2\left(-\frac{\sqrt{3}}{2} + i \cdot \frac{1}{2}\right) = -\sqrt{3} + i

k = 2: \theta_2 = \pi/6 + 4\pi/3 = 3\pi/2.

z_2 = 2\left(\cos\frac{3\pi}{2} + i\sin\frac{3\pi}{2}\right) = 2(0 + i(-1)) = -2i

Why: each root is at angle \pi/6 + k \cdot 2\pi/3 on a circle of radius 2. The three angles \pi/6, 5\pi/6, 3\pi/2 are 120° apart.

Step 4. Verify z_0^3 = 8i.

(\sqrt{3} + i)^2 = 3 + 2\sqrt{3}\,i + i^2 = 2 + 2\sqrt{3}\,i
(\sqrt{3} + i)^3 = (\sqrt{3} + i)(2 + 2\sqrt{3}\,i) = 2\sqrt{3} + 2 \cdot 3 \cdot i + 2i + 2\sqrt{3}\,i^2 = 2\sqrt{3} - 2\sqrt{3} + (6 + 2)i = 8i \;\checkmark

Why: a direct computation confirms the formula. The real parts cancel (2\sqrt{3} - 2\sqrt{3} = 0), leaving only the imaginary part 8i.

Result: The cube roots of 8i are \sqrt{3} + i, -\sqrt{3} + i, and -2i.

The three cube roots of 8i on the complex planeA complex plane with a circle of radius 2 centred at the origin. Three points are marked on this circle: root 3 plus i at angle pi over 6, negative root 3 plus i at angle 5 pi over 6, and negative 2i at angle 3 pi over 2. These three points form an equilateral triangle. The original number 8i is shown on the positive imaginary axis at distance 8. ReIm r = 2 π/6 √3 + i −√3 + i −2i 8i (original)
The three cube roots of $8i$ form an equilateral triangle on a circle of radius $2$. The first root $\sqrt{3} + i$ sits at angle $\pi/6 = 30°$. The other two are at $150°$ and $270°$, each $120°$ apart. The original number $8i$ is far up the imaginary axis — cubing any root multiplies the modulus to $8$ and triples the angle to reach $\pi/2$.

Notice that two of the three roots have the same imaginary part (+i), while the third root -2i is purely imaginary. None of the cube roots of 8i are real — which makes sense, because a real number cubed is real, not purely imaginary.

Example 2: Find the square roots of $-3 + 4i$

A complex number with no clean polar form — you work algebraically instead of using explicit trigonometric values.

Step 1. Suppose z = a + bi satisfies z^2 = -3 + 4i. Expand:

(a + bi)^2 = a^2 - b^2 + 2ab\,i

Why: for square roots, the direct algebraic approach is often cleaner than the polar form, since you get a system of two real equations.

Step 2. Match real and imaginary parts.

a^2 - b^2 = -3 \qquad \text{...(i)}
2ab = 4 \qquad \text{...(ii)}

From (ii): b = 2/a.

Why: two equations, two unknowns. From the imaginary equation, express b in terms of a and substitute into the real equation.

Step 3. Substitute into (i).

a^2 - \frac{4}{a^2} = -3

Multiply through by a^2:

a^4 + 3a^2 - 4 = 0

This is a quadratic in a^2. Let u = a^2:

u^2 + 3u - 4 = 0 \implies (u + 4)(u - 1) = 0

So u = 1 or u = -4. Since u = a^2 \ge 0, take u = 1, giving a = \pm 1.

Why: a is real, so a^2 cannot be negative. The root u = -4 is discarded.

Step 4. Find corresponding b values and write the roots.

If a = 1: b = 2/1 = 2. So z = 1 + 2i. If a = -1: b = 2/(-1) = -2. So z = -1 - 2i.

Why: the two square roots are negatives of each other, as always — if z^2 = w, then (-z)^2 = w too.

Step 5. Verify: (1 + 2i)^2 = 1 + 4i + 4i^2 = 1 + 4i - 4 = -3 + 4i. \checkmark

Result: The square roots of -3 + 4i are 1 + 2i and -1 - 2i.

The two square roots of negative 3 plus 4i shown on the complex planeA complex plane with three points marked: w equals negative 3 plus 4i in the second quadrant, and its two square roots 1 plus 2i in the first quadrant and negative 1 minus 2i in the third quadrant. The two roots are diametrically opposite on a circle of radius root 5 centred at the origin. A dashed line connects the two roots through the origin. ReIm 123 −1−2−3 12 −1−2 r = √5 w = −3 + 4i 1 + 2i −1 − 2i
The two square roots of $-3 + 4i$ are $1 + 2i$ and $-1 - 2i$, shown as red dots. They are diametrically opposite on a circle of radius $|z| = \sqrt{1^2 + 2^2} = \sqrt{5}$ (dashed circle). The original number $w = -3 + 4i$ sits at distance $|w| = \sqrt{9 + 16} = 5 = (\sqrt{5})^2$ from the origin, consistent with $|z|^2 = |w|$. Squaring either red dot lands at $w$.

The algebraic method (matching real and imaginary parts of z^2 = w) works cleanly whenever the resulting equation in a has rational or clean irrational roots. For higher powers, or when the algebra gets unwieldy, the polar form approach is more systematic. Both methods give the same answer.

Common confusions

Going deeper

If you can find the nth roots of any complex number using the polar form, understand why they form a regular polygon, and connect the general roots to the roots of unity, you have the core. The following material explores where these ideas lead.

The fundamental theorem of algebra

Every polynomial of degree n with complex coefficients has exactly n roots in \mathbb{C} (counted with multiplicity). The nth root formula is a special case: z^n - w = 0 is a degree-n polynomial, and it has exactly n roots. The fundamental theorem guarantees this works for every polynomial, not just binomials — but for general polynomials, there is no closed-form formula (when n \ge 5, by the Abel-Ruffini theorem).

Principal roots and branch cuts

When you write w^{1/n}, which of the n roots do you mean? The convention is to take the principal root: the one with the smallest positive argument (equivalently, k = 0 in the formula). But this choice is artificial — the map w \mapsto w^{1/n} cannot be made continuous on the entire complex plane. The points where continuity breaks form a branch cut, usually placed along the negative real axis. This is the beginning of the theory of multi-valued functions and Riemann surfaces — a topic central to complex analysis.

Solving z^n = w geometrically

Before the algebraic formula, there is a compass-and-straightedge question: can you construct the nth roots of a given complex number? The answer depends on n. You can always construct z_0 = R^{1/n} when R is constructible, and you can subdivide the angle \phi into n equal parts only for certain values of n. Angle trisection is famously impossible in general, which means the cube roots cannot always be constructed with compass and straightedge — one of the great results of classical geometry, settled only with the algebraic theory of field extensions.

Where this leads next