Chapter 3 Operations

In this section we will use the complex numbers \(z = a + bi\) and \(w = c + di\) to show how to do basic computation with complex numbers. Here, \(a\), \(b\), \(c\), and \(d\) are arbitrary real numbers.

3.1 Addition and subtraction

The addition and subtraction of complex numbers is just adding/subtracting together the real and imaginary components:

\(z + w = (a + bi) + (c + di) = a + c + bi + di = (a + c) + (b + d) i\)

\(z - w = (a + bi) - (c + di) = a - c + bi - di = (a - c) + (b - d) i\)

3.2 Multiplication

The multiplication of complex numbers is just multiplying the brackets out and then collecting the real and imaginary terms together:

\(z \cdot w = (a + bi) (c + di) = ac + adi + bci + bd (i^2) = ac + adi + bci - bd = (ac - bd) + (ad + bc) i\)

3.3 Complex conjugate, modulus and division

The complex conjugate of a number \(z\), denoted \(\bar{z}\) is obtained by flipping the sign of the imaginary part of \(z\).

If \[z = a + bi\] then, \[\bar{z} = a - bi .\]

Geometrically, taking the complex conjugate of a number is equivalent to reflecting/mirroring it over the real axis

The or of the complex number \(z = a + bi\), denoted \(|z|\) (or \(r\)).

Definition: \[| z | = \sqrt{a^2 + b^2}\]

Geometrically, the modulus is the straight line distance from the origin of the complex plane to the point representing the complex number (Pythagoras’ theorem! Since the modulus is the hypotenuse of a right angle triangle with sides \(a\) and \(b\)).

Note that multiplying a complex number by it’s conjugate always yields a real number:

\(z \cdot \bar{z} = (a + bi) (a - bi) = a^2 - abi + abi - b^2 i^2 = a^2 + b^2 = |z|^2\)

Dividing complex numbers is a bit more involved than the other arithmetic operations were. When we write \[\frac{z}{w} = \frac{a + bi}{c + di}\] , we notice that we have a sum in both the numerator and the denominator. We could solve this using partial fractions, but that would take too much time. Instead, we can multiply both parts of the fraction by the complex conjugate of the denominator. \[\frac{a + bi}{c + di} = \frac{(a + bi) (c - di)}{(c + di) (c - di)}\] since we’re multiplying both, the actual value of the fraction doesn’t change, but we did get rid of the sum in the denominator (which is now just a real number, equal to the modulus of \(c + di\)). If we multiply things out, we get \[\frac{(ac - bd) + (ad + bc) i}{c^2 + d^2} = \frac{ac - bd}{c^2 + d^2} + \frac{ad + bc}{c^2 + d^2} i\] Here’s an example to see it in practice as well: \[\frac{1 + 2 i}{2 + 3 i} = \frac{(1 + 2 i) (2 - 3 i)}{(2 + 3 i) (2 - 3 i)} = \frac{8 + i}{13} = \frac{8}{13} + \frac{1}{13} i\]