Lesson 1: An introduction to complex numbers

Introduction

In our study of mathematics, we find that the set of real numbers is not sufficient to solve all algebraic equations. For instance, consider the equation $x^2 = -1$. There is no real number that, when squared, results in a negative value.

To address this limitation, we must expand our concept of numbers. This leads us to the set of complex numbers. The first step in understanding this powerful new set is to become familiar with the foundational concept of the imaginary number, denoted by "i".

The imaginary number "i"

The imaginary number "i" is defined as the number whose square is -1.

i.e. $i^2 = -1$

Thus we can solve the equation $x^2 = -1$ as follows:

$$ \begin{flalign*} &\therefore \quad x^2 = -1 &&\\ &\therefore \quad x^2 = i^2 &&\\ &\therefore \quad x = \pm\sqrt{i^2} &&\\ &\therefore \quad x = \pm i &&\\ &\therefore \quad \text{The solution set} = \{i, -i\} && \end{flalign*} $$

Notice that

$i \times i = i^2 = -1$
$(-i) \times (-i) = i^2 = -1$

Remarks on Imaginary Numbers

The imaginary number "i" is not a real number. This means $i \notin \R$, so it cannot be plotted on the real number line.
Numbers like $3i$, $-2i$, and $\sqrt{5}i$ are examples of purely imaginary numbers.
For any positive real number 'a', the square root of its negative counterpart can be expressed as $\sqrt{-a} = \sqrt{a} \times i$. For example: $\sqrt{-2} = \sqrt{2}i$, $\sqrt{-3} = \sqrt{3}i$, and $\sqrt{-25} = 5i$.
Important: The product rule for square roots, $\sqrt{a} \times \sqrt{b} = \sqrt{ab}$, does not apply if both 'a' and 'b' are negative real numbers. $\sqrt{-1} \times \sqrt{-1} \neq \sqrt{(-1) \times (-1)}$
Because: $ \sqrt{-1} \times \sqrt{-1} = i \times i = i^2 = -1 $
But: $ \sqrt{(-1) \times (-1)} = \sqrt{1} = 1 $

Integer powers of "i"

The powers of "i" follow a repeating pattern. Since $i^2 = -1$, we can find the value of any integer power of "i":

$i^3 = i^2 \times i = -1 \times i = -i$
$i^4 = i^2 \times i^2 = -1 \times -1 = 1$
$i^5 = i^4 \times i = 1 \times i = i$
$i^6 = i^4 \times i^2 = 1 \times -1 = -1$

The integer powers of "i" always result in one of four values: $i, -1, -i, 1$. This cycle repeats every four powers.

Generally, for any integer $n \in \Z$:

$i^{4n} = (i^4)^n = 1^n = 1$
$i^{4n+1} = i^{4n} \times i^1 = 1 \times i = i$
$i^{4n+2} = i^{4n} \times i^2 = 1 \times -1 = -1$
$i^{4n+3} = i^{4n} \times i^3 = 1 \times -i = -i$

A useful shortcut

To find $i^n$ where n is an integer, we can find the remainder of the division $n \div 4$. If:

The remainder is 0, then $i^n = 1$
The remainder is 1, then $i^n = i^1 = i$
The remainder is 2, then $i^n = i^2 = -1$
The remainder is 3, then $i^n = i^3 = -i$

For example:

$i^{16} = 1$ (because $16 \div 4 = 4$ with remainder 0)
$i^{63} = -i$ (because $63 \div 4 = 15$ with remainder 3)
$i^{42} = -1$ (because $42 \div 4 = 10$ with remainder 2)
$i^{101} = i$ (because $101 \div 4 = 25$ with remainder 1)
$i^{4n+23} = -i$ (because $23 \div 4 = 5$ with remainder 3)

Properties of Powers of "i"

To simplify fractions like $\frac{1}{i^{19}}$, we can express "1" as a power of "i". The smallest multiple of 4 greater than 19 is 20. So, $1 = i^{20}$.
$\frac{1}{i^{19}} = \frac{i^{20}}{i^{19}} = i^{20-19} = i^1 = i$
The sum of any four consecutive integer powers of "i" is always zero. For any $n \in \Z$:
$i^n + i^{n+1} + i^{n+2} + i^{n+3} = 0$
For example: $i^6 + i^7 + i^8 + i^9$
$= (-1) + (-i) + (1) + (i) = 0$

The Complex Number

A complex number is a number that can be written in the form:

$a + bi$

Where 'a' and 'b' are real numbers, and $i^2 = -1$.

a is called the real part.
b is called the imaginary part.

Examples of complex numbers:

$2 - i$
$7 + 13i$
$5i - 4$ (or $-4 + 5i$)
$\sqrt{2} + \sqrt{3}i$

Remarks on Complex Numbers

If the imaginary part $b = 0$, the number $Z = a$ is a real number. For example, $Z=5$ is a real number.
If the real part $a = 0$ (and $b \neq 0$), the number $Z = bi$ is a purely imaginary number. For example, $Z=2i$ is an imaginary number.

From this, we can see that every real number is a complex number with an imaginary part of 0. Therefore, the set of real numbers ($\R$) is a subset of the set of complex numbers ($\C$).

The Set of Complex Numbers

The set of complex numbers, denoted by $\C$, is defined as: $$ \C = \{ a+bi : a \in \R, b \in \R, i^2 = -1 \} $$

Example 1

Find the solution set of each of the following equations in the set of complex numbers:

1) $2x^2 + 18 = 0$

$$ \begin{flalign*} &\therefore \quad 2x^2 + 18 = 0 &&\\ &\therefore \quad 2x^2 = -18 &&\\ &\therefore \quad x^2 = -9 &&\\ &\therefore \quad x = \pm\sqrt{-9} &&\\ &\therefore \quad x = \pm\sqrt{9i^2} &&\\ &\therefore \quad x = \pm 3i &&\\ &\therefore \quad \text{The solution set} = \{3i, -3i\} && \end{flalign*} $$

2) $x^2 + x + 1 = 0$

Using the quadratic formula where $a=1, b=1, c=1$:

$$ \begin{flalign*} &\therefore \quad x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} &&\\ &\therefore \quad x = \frac{-1 \pm \sqrt{1^2 - 4 \times 1 \times 1}}{2 \times 1} &&\\ &\therefore \quad x = \frac{-1 \pm \sqrt{-3}}{2} &&\\ &\therefore \quad x = \frac{-1 \pm \sqrt{3}i}{2} &&\\ &\therefore \quad x = -\frac{1}{2} \pm \frac{\sqrt{3}}{2}i &&\\ &\therefore \quad \text{The solution set} = \{-\frac{1}{2} + \frac{\sqrt{3}}{2}i, -\frac{1}{2} - \frac{\sqrt{3}}{2}i\} && \end{flalign*} $$

Equality of two complex numbers

Two complex numbers are equal if and only if the two real parts are equal and the two imaginary parts are equal.

i.e. If $(a+bi)$ and $(c+di)$ are two complex numbers and if $a=c, b=d$, then $a+bi = c+di$.
and vice versa If $a+bi = c+di$, then $a=c, b=d$.

Notice that Order in complex numbers whose imaginary part not equal to zero has no meaning, we do not know which is greater $(5+3i)$ or $(-4+7i)$?

Example 2

Find the values of x and y which satisfy each of the following where $x \in \R, y \in \R, i^2 = -1$:

1) $(2x-3) + 5i = 7 + (3-2y)i$

$$ \begin{flalign*} &\text{By equating the real parts:}&&\\ &\therefore \quad 2x - 3 = 7 &&\\ &\therefore \quad 2x = 10 &&\\ &\therefore \quad x = 5 &&\\ &\text{By equating the imaginary parts:}&&\\ &\therefore \quad 3 - 2y = 5 &&\\ &\therefore \quad -2y = 2 &&\\ &\therefore \quad y = -1 && \end{flalign*} $$

2) $x+yi = \sqrt{-4} + i^{22}$

$$ \begin{flalign*} &\therefore \quad x + yi = 2i + i^{4(5)+2} &&\\ &\therefore \quad x + yi = 2i + i^2 &&\\ &\therefore \quad x + yi = 2i + (-1) &&\\ &\therefore \quad x + yi = -1 + 2i &&\\ &\therefore \quad x = -1 \text{ , } y = 2 && \end{flalign*} $$

3) $x-3y+(2x+y)i = 6+5i$

$x - 3y = 6 \qquad \cdots(1)$

$2x + y = 5 \qquad \cdots(2)$

Multiply the equation (2) by 3

$\therefore \quad 6x + 3y = 15 \qquad \cdots(3)$

By adding (1) and (3):

$\therefore \quad 7x = 21$

$\therefore \quad x = 3$

By substituting in (2):

$\therefore \quad y = -1$

Adding and subtracting complex numbers

When adding or subtracting two complex numbers, we add or subtract real parts together and add or subtract imaginary parts together.

Example 3

Find the result of each of the following in the simplest form:

1) $(3+7i^{13}) + (5-9i)$

$\therefore i^{13} = i$

$\therefore \text{The expression} = (3+7i)+(5-9i)$

$= (3+5) + (7i-9i)$

$= 8-2i$

2) $(2-\sqrt{-16}) - (5-i)$

$\therefore \sqrt{-16} = 4i$

$\therefore \text{The expression} = (2-4i) - (5-i)$

$= (2-4i) + (-5+i)$

$= (2-5) + (-4i+i)$

$= -3-3i$

Multiplying complex numbers

Find the result of each of the following in the simplest form:

1) $(4+3i)(2-5i)$

$= 4(2-5i) + 3i(2-5i)$

$= 8 - 20i + 6i - 15i^2$

$= 8 - 20i + 6i + 15$ (where $i^2 = -1$)

$= (8+15) + (-20i+6i)$

$= 23 - 14i$

2) $(5-2i)(5+2i)$

$= 25 - 4i^2$

$= 25 + 4$ (where $i^2 = -1$)

$= 29$

Remember that $(a+b)(a-b) = a^2 - b^2$

3) $(3+2i)^2$

$= 9 + 12i + 4i^2$

$= 9 + 12i - 4$ (where $i^2 = -1$)

$= 5 + 12i$

Remember that $(a \pm b)^2 = a^2 \pm 2ab + b^2$

4) $(1-i)^4$

$= ((1-i)^2)^2$

$= (1 - 2i + i^2)^2$

$= (1 - 2i - 1)^2$

$= (-2i)^2$

$= 4i^2$

$= -4$

Remark

$$(1 \pm i)^{2n} = (\pm 2i)^n \text{ where } n \in \Z$$

Proof: $(1 \pm i)^{2n} = [(1 \pm i)^2]^n$

$= [1 \pm 2i - 1]^n$

$= (\pm 2i)^n$

This remark is used to simplify some complex numbers as the following:

1) $(1+i)^{200}$

$= (2i)^{100}$

$= 2^{100}i^{100}$

$= 2^{100}$

2) $(3-3i)^4$

$= 3^4(1-i)^4$

$= 3^4(-2i)^2$

$= 3^4 \times 2^2 i^2$

$= -324$

The two conjugate numbers

The two numbers $a+bi$ and $a-bi$ are called conjugate numbers.

Note: Take care that the complex number and its conjugate differ only in the sign of their imaginary parts.

For example: The two numbers $3+4i$ , $3-4i$ are conjugate numbers.

Remarks

The conjugate of the number $2i-5$ is the number $-2i-5$ not $2i+5$
The conjugate of the number $2i$ is $-2i$
The conjugate of the number 3 is 3
The sum of the two conjugate numbers is always a real number, and the product of the two conjugate numbers is always a real number. For example The complex number $3+4i$ its conjugate is $3-4i$, then:
* Their sum = $(3+4i)+(3-4i)$
$= (3+3)+(4i-4i)$
$= 6 \in \R$
* Their product = $(3+4i)(3-4i)$
$= 9 - 16i^2$
$= 9+16$
$= 25 \in \R$

Simplify to the simplest form

Notice: To simplify the fraction whose denominator is a complex number, we multiply its two terms by the conjugate of denominator.

1) $\frac{4-3i}{i}$

$ = \frac{4-3i}{i} \times \frac{-i}{-i} $

$ = \frac{-4i+3i^2}{-i^2} $

$ = \frac{-4i-3}{-(-1)} $

$ = -3-4i $

2) $\frac{10}{3+i}$

$\therefore$ The conjugate of the denominator is $(3-i)$

$ = \frac{10}{3+i} \times \frac{3-i}{3-i} $

$ = \frac{10(3-i)}{9-i^2} $

$ = \frac{10(3-i)}{9+1} $

$ = \frac{10(3-i)}{10}$

$= 3-i $

3) $\frac{3+2i}{2-5i}$

$ = \frac{(3+2i)(2+5i)}{(2-5i)(2+5i)} $

$ = \frac{6+15i+4i+10i^2}{4-25i^2} $

$ = \frac{6+19i-10}{4+25} $

$ = \frac{-4+19i}{29} $

$ = -\frac{4}{29} + \frac{19}{29}i $

4) $\frac{(2+i)(1-i)}{(1+i)(3-2i)}$

$ = \frac{2-2i+i-i^2}{3-2i+3i-2i^2} $

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

$ = \frac{3-i}{5+i} $

$ = \frac{(3-i)(5-i)}{(5+i)(5-i)} $

$ = \frac{15-3i-5i+i^2}{25-i^2} $

$ = \frac{14-8i}{26} $

$ = \frac{2(7-4i)}{26} $

$ = \frac{7-4i}{13}$

$= \frac{7}{13} - \frac{4}{13}i $

Example 6

If $x = \frac{7-i}{2-i}$ and $y = \frac{13-i}{4+i}$

Prove that: x and y are conjugate numbers,

then prove that: $x^2+y^2=16$

$\therefore x = \frac{7-i}{2-i}$

$= \frac{(7-i)(2+i)}{(2-i)(2+i)}$

$ = \frac{14+7i-2i-i^2}{4-i^2} $

$ = \frac{14+5i+1}{4+1} $

$ = \frac{15+5i}{5}$

$= 3+i $

$, y = \frac{13-i}{4+i}$

$= \frac{(13-i)(4-i)}{(4+i)(4-i)}$

$ = \frac{52-13i-4i+i^2}{16-i^2} $

$ = \frac{52-17i-1}{16+1} $

$ = \frac{51-17i}{17}$

$= 3-i $

$\therefore$ x and y are conjugate numbers "Notice that the signs of the imaginary parts are different."

$, x^2 = (3+i)^2$

$ = 9+6i+i^2 $

$ = 8+6i $

$, y^2 = (3-i)^2$

$ = 9-6i+i^2 $

$ = 8-6i $

$\therefore x^2+y^2$

$ = (8+6i)+(8-6i) $

$ = (8+8)+(6i-6i) $

$ = 16 $

Introduction

The imaginary number "i"

Notice that

Remarks on Imaginary Numbers

Integer powers of "i"

A useful shortcut

For example:

Properties of Powers of "i"

The Complex Number

Examples of complex numbers:

Remarks on Complex Numbers

The Set of Complex Numbers

Example 1

1) \(2x^2 + 18 = 0\)

2) \(x^2 + x + 1 = 0\)

Equality of two complex numbers

Example 2

1) \((2x-3) + 5i = 7 + (3-2y)i\)

2) \(x+yi = \sqrt{-4} + i^{22}\)

3) \(x-3y+(2x+y)i = 6+5i\)

Adding and subtracting complex numbers

Example 3

1) \((3+7i^{13}) + (5-9i)\)

2) \((2-\sqrt{-16}) - (5-i)\)

Multiplying complex numbers

1) \((4+3i)(2-5i)\)

2) \((5-2i)(5+2i)\)

3) \((3+2i)^2\)

4) \((1-i)^4\)

Remark

1) \((1+i)^{200}\)

2) \((3-3i)^4\)

The two conjugate numbers

Remarks

Simplify to the simplest form

1) \(\frac{4-3i}{i}\)

2) \(\frac{10}{3+i}\)

3) \(\frac{3+2i}{2-5i}\)

4) \(\frac{(2+i)(1-i)}{(1+i)(3-2i)}\)

Example 6