Complex Number Calculator

Result
\frac{1}{2} + \frac{3}{2}i

In the form \displaystyle a + bi: \displaystyle \frac{1}{2} + \frac{3}{2}i.

  • Type: Complex numbers
  • Method: Multiply by the conjugate
  • Real part: \displaystyle \frac{1}{2}
  • Imaginary part: \displaystyle \frac{3}{2}
  • Conjugate: \displaystyle \frac{1}{2} - \frac{3}{2}i
  • Modulus: \displaystyle |z| = \sqrt{\frac{5}{2}} = \frac{\sqrt{10}}{2}
  • Argument: \displaystyle 71.57^\circ
  • Also: Expand it

Step-by-step solution

The problem

\frac{2 + i}{1 - i}
1

Multiply top and bottom by the conjugate \displaystyle 1 + i

\frac{2 + i}{1 - i} \cdot \frac{1 + i}{1 + i}

The conjugate changes the sign of the imaginary part; \displaystyle (a + bi)(a - bi) = a^2 + b^2 is a real number, so the \displaystyle i leaves the denominator.

2

Multiply out the top and the bottom

\begin{aligned} \text{top: } \left(2 + i\right)\left(1 + i\right) &= i^{2} + i + 2i + 2 = -1 + i + 2i + 2 = 1 + 3i \\[4pt] \text{bottom: } 1^{2} + \left(-1\right)^{2} &= 2 \end{aligned}

Use \displaystyle i^2 = -1.

3

Divide each part by \displaystyle 2

= \frac{1 + 3i}{2} = \frac{1}{2} + \frac{3}{2}i

Write the answer as \displaystyle a + bi.

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

Try an example

Arithmetic (2 + 3i)(1 - i)(2 + i)/(1 - i)i^2025
Polar form polar form of 1 + ipolar form of -1 + sqrt(3) i
Powers (De Moivre) (1 + i)^8(1 + sqrt(3) i)^6
Roots cube roots of 8fourth roots of -16square roots of i

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.