Introduction to Complex Numbers

Suppose we wanted to take the square √ of -1. We know that there aren't any real numbers that would give a suitable answer so mathematicians introduced a number "i" known as the imaginary unit.

i² = -1

√-1 = i

√-4 = 2i

√ -3 = i√3

These numbers are known as imaginary numbers. They are written in the form ib (where b is a real number. b is placed in front of i if it is a rational number, otherwise it is placed on the back)

A complex number is a number that has both a real and a imaginary component. They are written in the form a+ ib (where a and b are real numbers)

All numbers are complex numbers. Real numbers are simply complex numbers with a zero imaginary component. Eg 1 = 1+ 0i. Imaginary numbers are complex numbers with a zero real component. Eg. 4i = 0 +4i

Equal complex numbers

Two complex numbers are equal only if their real components are equal and their imaginary components are also equal.

Eg. (a + ib) = (c + id) iff a = c and b = d.

Fyi. Iff is a mathematical term which means if and only if.

An example question can be found here.


Quadratic with negative discriminant

With complex numbers we can solve any quadratic equation even if the discriminant is less than 0.

Eg.  x² + x + 1 =0

x = (-1 ± √-3)/2

x = -1/2 + i√3/2 or 1/2 - i√3/2

Conjugate of a complex number

If z = a + ib the conjugate of z, written as

= a – ib

z= (a+ ib) (a –ib) = a² – i²b²

=a² + b²

The conjugate of a complex number is used to change the denominator into a real number if it is a complex number.

Eg. 2/1-2i = 2 (1+2i)/(1-2i)(1+2i) = (2 + 4i)/5

Sum, difference, product and division of complex numbers.

You do not have to remember these as they can be easily worked out with actual numbers.

Sum:

(a+ib) + (c+id) = (a+c) + i(b+d)

Difference:

(a+ib) – (c+id) = (a-c) + i(b-d)

Product:

(a+ib)(c+id) = ac + iad + ibc + i²bd

=ac – bd + i(ad+bc)

Division:

(a+ib)/(c+id) = (a+ib)(c-id)/(c+id)(c-id) (multiply by conjugate)

= (ac –iad +ibc –i²bd)/(c² +d²)

= [ac + bd + i(bc – ad)]/(c² +d²)

Example questions

Let z = 5 –i, find:

i)           z² in the form of a+ib

ii)          z + 2 in the form of a+ib

iii)         i/z in the form of a+ib

Solution

i)      z² = (5-i)²

= 25 – 10i + i²

= 24 – 10i

ii)      = 5 +i

z + 2 = (5-i) + 2(5+i)

= 5 – i + 10 +2i

= 15 + i

iii)     i/z = i/(5-i)

= i(5+i)/(5-i)(5+i)

= (-1 + 5i)/26

= -1/26 + 5/26 i