De Moivre’s Theorem Calculator (Powers and nth Roots of Complex Numbers)
Raise a complex number a + bi to the nth power or find all n of its nth roots with De Moivre’s theorem: modulus, argument, polar form and every root in a + bi form.
How it works
z = r(cos θ + i sin θ) · zⁿ = rⁿ(cos nθ + i sin nθ) · nth roots: w_k = r^(1/n)[cos((θ + 2πk)/n) + i sin((θ + 2πk)/n)], k = 0, 1, …, n − 1
Writing z in polar form with modulus r = √(a² + b²) and argument θ, De Moivre’s theorem says that raising to the nth power raises the modulus to the nth power and multiplies the angle by n. Reversing it gives exactly n distinct nth roots: all have modulus r^(1/n) and they are spaced 360°/n apart around a circle, forming a regular polygon.
Worked example
Cube roots of z = −8: r = 8 and θ = 180°, so every root has modulus 8^(1/3) = 2 at angles 60°, 180° and 300°. That gives 2(cos 60° + i sin 60°) = 1 + 1.732051i, 2(cos 180° + i sin 180°) = −2 and 1 − 1.732051i. Roots: 1 + 1.732051i; −2; 1 − 1.732051i.
Sources: OpenStax Precalculus 2e — 8.5 Polar Form of Complex Numbers · Wolfram MathWorld — de Moivre’s Identity
Frequently asked questions
How many nth roots does a complex number have?
Exactly n (for z ≠ 0). A number has 2 square roots, 3 cube roots, 4 fourth roots and so on, evenly spaced on a circle.
How do I convert a + bi to polar form?
r = √(a² + b²) and θ = atan2(b, a), the angle in the correct quadrant. Then z = r(cos θ + i sin θ), often written r·cis θ or r·e^(iθ).
Why are −8’s cube roots not just −2?
−2 is the only real cube root, but in the complex numbers there are two more, 1 ± √3·i, because rotating by 120° keeps the cube equal to −8.
How do I compute (1 + i)¹⁰ quickly?
r = √2 and θ = 45°, so (1 + i)¹⁰ = (√2)¹⁰(cos 450° + i sin 450°) = 32(cos 90° + i sin 90°) = 32i.
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This calculator is for educational purposes. Double-check important results before acting on them.