Complex Number Calculator

Result
2,\quad -1 + \sqrt{3}i,\quad -1 - \sqrt{3}i

The 3 cube roots of \displaystyle 8.

  • Type: Complex roots
  • Method: De Moivre's theorem

Step-by-step solution

The problem

\sqrt[3]{8}
1

Write \displaystyle 8 in polar form

8 = 8\left(\cos 0 + i\sin 0\right)

Modulus \displaystyle r = 8, argument \displaystyle \theta = 0.

2

Use the root formula

w_k = 2\left(\cos\frac{0 + 2\pi k}{3} + i\sin\frac{0 + 2\pi k}{3}\right)

The 3 roots have modulus \displaystyle \sqrt[3]{r} and arguments \displaystyle \frac{\theta + 2\pi k}{3} for \displaystyle k = 0, 1, \dots, 2 — equally spaced around a circle.

3

List the roots

\begin{aligned} k &= 0:\quad \theta = 0,\quad w_0 = 2 \\[4pt] k &= 1:\quad \theta = \frac{2\pi}{3},\quad w_1 = -1 + \sqrt{3}i \\[4pt] k &= 2:\quad \theta = \frac{4\pi}{3},\quad w_2 = -1 - \sqrt{3}i \end{aligned}

Use i for √−1: (2 + i)/(1 - i), polar form of 1 + i, (1 + i)^8, cube roots of 8i.

Add, subtract, multiply and divide complex numbers, or convert them to polar form, with each line of working shown – for example (2 + 3i)(1 − i) = 5 + i, using i² = −1. It also finds the modulus and argument, powers by De Moivre’s theorem and every nth root.

Arithmetic with complex numbers

  • Add and subtract the real parts and the imaginary parts separately.
  • Multiply like brackets, then replace i^2 by -1.
  • Divide by multiplying the top and bottom by the conjugate of the bottom: (a + bi)(a - bi) = a^2 + b^2 is real.
  • Powers of i repeat every four: i, -1, -i, 1.

Example: divide 3 + 2i by 1 - i – multiply top and bottom by the conjugate 1 + i:

\frac{3 + 2i}{1 - i} \cdot \frac{1 + i}{1 + i} = \frac{3 + 3i + 2i + 2i^2}{1 - i^2} = \frac{1 + 5i}{2} = \frac{1}{2} + \frac{5}{2}i

Check by multiplying back: \left(\tfrac{1}{2} + \tfrac{5}{2}i\right)(1 - i) = 3 + 2i. Open it in the calculator.

Any power of i in one step: divide the power by 4 and keep the remainder – 0, 1, 2, 3 give 1, i, −1, −i. So i²⁰²⁶ = −1, because 2026 leaves a remainder of 2.

Polar form

A complex number z = a + bi is also a point at distance r from the origin and angle \theta:

z = r(\cos\theta + i\sin\theta) = re^{i\theta},\qquad r = \sqrt{a^2 + b^2},\quad \cos\theta = \frac{a}{r},\ \sin\theta = \frac{b}{r}

Watch the quadrant. The angle is not simply \tan^{-1}(b/a): for -1 - i that gives 45°, but the point is in the third quadrant, so the argument is −3π/4 (−135°, or 225° measured the long way round). The calculator picks the angle from the signs of a and b, as the step shows.

De Moivre’s theorem and roots

\left[r(\cos\theta + i\sin\theta)\right]^n = r^n(\cos n\theta + i\sin n\theta)

The n nth roots of re^{i\theta} all have modulus \sqrt[n]{r} and arguments \frac{\theta + 2\pi k}{n} for k = 0, 1, \dots, n - 1 – evenly spaced around a circle.

Example: (1 + i)^8

1 + i = \sqrt{2}\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right) \;\Rightarrow\; (1 + i)^8 = (\sqrt{2})^8(\cos 2\pi + i\sin 2\pi) = 16

Where complex numbers are used

  • AC electricity: impedance is a complex number, Z = R + jX (engineers write j, since i means current). A 30 Ω resistor in series with a 40 Ω reactance has |Z| = 50 Ω, the same 3-4-5 arithmetic as the modulus.
  • Rotations: multiplying by i turns a point a quarter turn about 0; multiplying by e^{i\theta} turns it by θ – the reason polar form makes powers easy.
  • Signals and waves: sound, radio and the Fourier transform describe each frequency with a complex number – its size is the strength, its angle the timing (phase).

Quadratics with negative discriminants give complex solutions too – try the quadratic equation solver.

Frequently asked questions

What is i?

The imaginary unit, defined by i² = −1. Complex numbers are written a + bi, with a real part a and an imaginary part b.

What are the modulus and argument?

The modulus r = √(a² + b²) is the distance from 0; the argument θ is the angle from the positive real axis, found from cos θ = a/r and sin θ = b/r.

How many cube roots does a number have?

Every non-zero complex number has exactly three cube roots (and n nth roots). For 8 they are 2, −1 + √3 i and −1 − √3 i.