Algebraic Form of Complex Numbers and Basic Arithmetic Operations
A complex number in algebraic form is z = a + bi, where a and b are real numbers and i is the imaginary unit defined by i2 = −1; a is the real part and b is the imaginary part. In brief: (3 + 2i) + (1 − 5i) = 4 − 3i and (3 + 2i)(1 − 5i) = 3 − 15i + 2i − 10i2 = 13 − 13i. To divide, multiply the numerator and the denominator by the conjugate of the denominator.
The real numbers are not enough to solve every polynomial equation: x2 + 1 = 0 has no real solution. Complex numbers extend the real numbers so that every polynomial equation has a solution. They are used in electrical engineering, quantum mechanics and signal processing. This guide explains the algebraic form, the four operations, powers of i and the conjugate, with a table of rules and worked examples.
What is a complex number?
A complex number is z = a + bi with real a and b. We write Re z = a and Im z = b (the imaginary part is the real number b, not bi). If b = 0, z is a real number; if a = 0 and b ≠ 0, z is purely imaginary. Two complex numbers are equal if and only if both their real parts and their imaginary parts are equal: a + bi = c + di means a = c and b = d.
The set of complex numbers ℂ contains the real numbers ℝ. The square root of a negative real number is purely imaginary: √(−9) = 3i, since (3i)2 = −9. A caution: √(−a)√(−b) = −√(ab) for positive a and b, not +√(ab): for example √(−4) · √(−9) = 2i · 3i = −6.
How do you add, subtract, multiply and divide complex numbers?
Addition and subtraction: combine the real and imaginary parts separately: (a + bi) ± (c + di) = (a ± c) + (b ± d)i.
Multiplication: multiply like binomials and use i2 = −1: (a + bi)(c + di) = (ac − bd) + (ad + bc)i.
Division: multiply the numerator and the denominator by the conjugate of the denominator. The conjugate of c + di is c − di, and (c + di)(c − di) = c2 + d2, which is real. So (a + bi)/(c + di) = (a + bi)(c − di)/(c2 + d2).
Example 1
(5 + 3i) − (2 − 4i) = 3 + 7i. And 2(1 + i) + 3(2 − i) = 2 + 2i + 6 − 3i = 8 − i.
Example 2
(2 + 3i)(4 − i) = 8 − 2i + 12i − 3i2 = 8 + 10i + 3 = 11 + 10i. And (1 + i)2 = 1 + 2i + i2 = 2i.
Example 3
(3 + 4i)/(1 − 2i) = (3 + 4i)(1 + 2i)/((1 − 2i)(1 + 2i)) = (3 + 6i + 4i + 8i2)/5 = (3 − 8 + 10i)/5 = −1 + 2i. Check: (−1 + 2i)(1 − 2i) = −1 + 2i + 2i − 4i2 = 3 + 4i ✓.
Example 4: reciprocal
1/i = i/i2 = −i. And 1/(2 + i) = (2 − i)/5 = 0.4 − 0.2i.
What are the powers of i?
The powers repeat with period 4: i1 = i, i2 = −1, i3 = −i, i4 = 1, i5 = i, and so on. To compute in, divide n by 4 and use the remainder: remainder 0 gives 1, 1 gives i, 2 gives −1 and 3 gives −i.
Example 5
i27: 27 = 4 · 6 + 3, so i27 = i3 = −i. And i100 = 1. And i−1 = −i. And the sum i + i2 + i3 + i4 = 0, so any four consecutive powers add up to zero.
How do you solve equations with complex numbers?
Example 6: a quadratic with negative discriminant
x2 + 4x + 13 = 0. The discriminant is 16 − 52 = −36, so x = (−4 ± 6i)/2 = −2 ± 3i. Check: (−2 + 3i)2 = 4 − 12i − 9 = −5 − 12i; and 4(−2 + 3i) = −8 + 12i; sum: −5 − 12i − 8 + 12i + 13 = 0 ✓.
Example 7: a linear equation
(2 + i)z = 5 − 5i. Then z = (5 − 5i)/(2 + i) = (5 − 5i)(2 − i)/5 = (10 − 5i − 10i + 5i2)/5 = (5 − 15i)/5 = 1 − 3i.
Example 8: comparing parts
Find real x and y with (x + 2i)(3 − yi) = 8 + 4i. Expand: 3x − xyi + 6i − 2yi2 = (3x + 2y) + (6 − xy)i. Equating the real and imaginary parts gives 3x + 2y = 8 and 6 − xy = 4, so xy = 2. Substitute y = 2/x: 3x + 4/x = 8, so 3x2 − 8x + 4 = 0, with D = 64 − 48 = 16 and x = (8 ± 4)/6, which gives x = 2 or x = 2/3. The solutions are (x, y) = (2, 1) and (2/3, 3). Check the second: (2/3 + 2i)(3 − 3i) = 2 − 2i + 6i − 6i2 = 8 + 4i ✓.
Table of operations and powers of i
| Operation | Rule | Example: z = 2 + 3i, w = 1 − i |
|---|---|---|
| z + w | (a + c) + (b + d)i | 3 + 2i |
| z − w | (a − c) + (b − d)i | 1 + 4i |
| zw | (ac − bd) + (ad + bc)i | (2 + 3) + (−2 + 3)i = 5 + i |
| z/w | z · conj(w)/(c2 + d2) | (2 + 3i)(1 + i)/2 = (−1 + 5i)/2 |
| conjugate of z | a − bi | 2 − 3i |
| z · conj(z) | a2 + b2 | 13 |
| n mod 4 | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| in | 1 | i | −1 | −i |
Where do complex numbers come from?
Gerolamo Cardano met square roots of negative numbers in the sixteenth century while solving cubic equations, and Rafael Bombelli described rules for calculating with them in 1572. Descartes coined the term “imaginary” as a dismissive label. Euler introduced the symbol i in the eighteenth century, and Gauss gave the geometric representation and the name “complex numbers” in the early nineteenth century, which made them widely accepted.
Where are they used today?
In electrical engineering, voltages and currents in AC circuits are complex numbers: an impedance Z = R + iX combines resistance R and reactance X. Quantum mechanics describes states with complex amplitudes. Signal processing uses the Fourier transform, which is built on complex exponentials, and computer graphics uses them for rotations in the plane.
Which mistakes are common?
Forgetting that i2 = −1
In multiplication, replace i2 by −1 and collect the real parts.
Writing the imaginary part as bi
For z = 3 + 4i, Im z = 4, not 4i.
Dividing without the conjugate
Do not leave i in the denominator. Multiply by the conjugate.
Using √(ab) = √a √b for negative numbers
√(−4) · √(−9) = −6, while √36 = 6. The rule holds only when at least one of the numbers is nonnegative.