Complex Root Calculator | Find Roots of Complex Numbers

Calculate the nth roots of any complex number. Find square roots, cube roots, and higher-order roots with real and imaginary components.

Find all nth roots of any complex number with our free calculator. Enter real and imaginary parts and the root order to compute every root using De Moivre's theorem.

Key Takeaway

Use the free Complex Root Calculator to calculate the nth roots of any complex number. find square roots, cube roots, and higher-order roots with real and imaginary components. Get instant results with step-by-step explanations.

How to Use the Complex Root Calculator

Enter the real part (a) of the complex number a + bi.
Enter the imaginary part (b) of the complex number a + bi.
Select the root order n (2 for square root, 3 for cube root, etc.).
Review all n roots with their real and imaginary components.

The Formula

The nth roots of z = r(cos θ + i sin θ) are r^(1/n)(cos((θ + 2kπ)/n) + i sin((θ + 2kπ)/n)) for k = 0, 1, ..., n-1.

Variable Definitions

r: Modulus of the complex number: √(a² + b²)
θ: Argument (angle) of the complex number: atan2(b, a)
n: The root order (2 = square root, 3 = cube root, etc.)
k: Root index from 0 to n-1, giving each distinct root
i: The imaginary unit where i² = -1

Finding the Cube Roots of 8i

Compute all 3 cube roots of z = 8i.

Step 1: Convert to polar: r = |8i| = 8, θ = π/2 (since 8i is on the positive imaginary axis).
Step 2: Compute r^(1/3) = 8^(1/3) = 2.
Step 3: For k = 0: angle = (π/2 + 0)/3 = π/6. Root = 2(cos(π/6) + i sin(π/6)) = √3 + i.
Step 4: For k = 1: angle = (π/2 + 2π)/3 = 5π/6. Root = 2(cos(5π/6) + i sin(5π/6)) = -√3 + i.
Step 5: For k = 2: angle = (π/2 + 4π)/3 = 3π/2. Root = 2(cos(3π/2) + i sin(3π/2)) = -2i.

Frequently Asked Questions

How many nth roots does a complex number have?

A nonzero complex number has exactly n distinct nth roots. They are evenly spaced around a circle of radius r^(1/n) in the complex plane, separated by angles of 2π/n.

What is De Moivre's theorem?

De Moivre's theorem states that for any complex number in polar form z = r(cos θ + i sin θ), z^n = r^n(cos(nθ) + i sin(nθ)). It extends to roots as well.

How do I find the square root of a complex number?

For z = a + bi, the two square roots are ±(√((r+a)/2) + i·sign(b)·√((r-a)/2)), where r = |z| = √(a² + b²).

Why are complex roots important?

Complex roots are essential in polynomial theory (Fundamental Theorem of Algebra), signal processing, electrical engineering, quantum mechanics, and solving differential equations.

Can I find roots of negative real numbers?

Yes. For example, √(-1) = i, and the cube roots of -8 are -2, 1 + i√3, and 1 - i√3. Any negative real number has complex nth roots when n is even.

What is the principal root?

The principal root is typically the root with the smallest non-negative argument (angle). For square roots, it is the one in the upper half-plane or on the positive real axis.

How are complex roots distributed?

All n roots lie on a circle of radius r^(1/n) centered at the origin. They are equally spaced at angles 2π/n apart, forming a regular n-gon in the complex plane.

What is the relationship between roots and conjugates?

If z is a root of a polynomial with real coefficients, then z̄ (its conjugate) is also a root. Complex roots always come in conjugate pairs for real-coefficient polynomials.

Complex Root Calculator – Guide & Formulas