**Complex numbers** arise naturally when solving quadratic equations. We know that the solutions to \( x^2 -1 =0 \) are \( x = 1, -1 \). What are the solutions to \( x^2 + 1 = 0 \)?

\( i \) is often used to denote the imaginary unit, which satisfies the equation \( i^2 = -1 \). \( i \) and \( -i \) will be the roots to the equation \( x^2 + 1 = 0 \). With this symbol, we can extend the real numbers to obtain the set of complex numbers, which are of the form \( z= a+ bi \), where \( a \) and \( b \) are real numbers.

We say that for a complex number \( z= a+bi \), it has real part \( a \), denoted as \( Re(z) \), and imaginary part \( b \), denoted as \( Im (z) \). Take note that the imaginary part is a real number. Let's see how the usual arithmetic operations work:

1) Addition:

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

2) Subtraction:

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

3) Multiplication:

\( (a+bi)\times (c+di) = ac + adi + bci + bdi^2 = (ac-bd) + (ad+bc) i \).

Division becomes slightly tricky, because we only know how to divide by a real number. Rewriting \( \frac {a+bi} { c+di}\) as \( \frac {a}{c+di} + \frac {bi}{c+di} \) isn't very helpful. If only we could make the denominator a real number ... To do so, we introduce the idea of a conjugate:

The

conjugateof the complex number \( z = a+bi \) is \( \overline{z} = a - bi \). Notice that the conjugate of the conjugate is the identity, i.e. \( \overline{ \overline{z}} = z \).

We now have:

\( z \overline{z} = (a+bi)(a-bi) = [ a \times a - b \times (-b) ] + [a \times (-b) + b \times a]i = a^2 + b^2 \)

using our nifty multiplication formula. This gives us a real non-negative value. Now, we introduce the idea of a Norm and absolute value:

The

normof the complex number \( z = a+bi \) is \( N( z ) = z \overline{z} = a^2 + b^2 \).The absolute value of the complex number is the positive square root of the norm, and is given by \( \left| a+bi \right| = \sqrt{a^2+b^2}\).

With this, we have the following:

4) Division: If \( c+di \) is non-zero, then \( \frac {a+bi} {c+di} = \frac {(a+bi)(c-di)}{(c+di)(c-di)} = \frac {(ac-bd) + (-ad+bc)i}{c^2 + d^2 } \).

\( z = x + iy \) is known as the **Rectangular Form** of the complex number. In an upcoming post, we will study the Polar Form of complex numbers.

## Worked Examples

## 1. If \( a \) and \( b \) are real numbers such that \( a + bi = 0 \), then \( a = 0 \) and \( b = 0 \).

This should be obvious to you, but let's show it. If \( a + bi = 0 \) , then \( a = - bi \). Squaring both sides, we get \( a^2 = (-bi)^2 = b^2 i^2 = - b^2 \). By non-negativity of squares, we have \( 0 \leq a^2 = -b^2 \leq 0 \), which implies that equality must hold throughout. Thus, \( 0 = a^2 = -b^2 \), which gives \( a = 0, b= 0 \).

Corollary: This allows us to compare coefficients of real and imaginary parts. In particular, if \( z = w \), then we must have \( Re(z) = Re(w), Im(z) = Im(w) \).

## 2. With real numbers, we are familiar with the concept of reciprocals. For example, 2 and \( \frac {1}{2} \) are reciprocals of each other because \( 2 \times \frac {1}{2} = 1 \). What is the reciprocal of a non-zero complex number \( z = a+ bi \)?

Solution 1: Since \( (a, b) \neq (0, 0) \), \( \left \| z \right \| \neq 0 \). We seek the value of \( \frac {1}{z} \) so will use the division operation, to obtain that \( \frac {1}{z} = \frac {1}{a+bi} = \frac {a-bi}{(a+bi) (a-bi)} = \frac {a-bi} {a^2 + b^2 } \).

Solution 2: Using the language of norm and conjugates, we can express this directly as:\( \frac {1}{z} = \frac {\overline{z} } { z \overline{z} } = \frac {\overline{z} } { N( z ) } \).

## 3. Verify that conjugation distributes over multiplication and division. Specifically, show that \( \overline{z \times w} = \overline{z} \times \overline{w} \) and \( \overline{ \left( \frac {z}{w} \right)} = \frac {\overline{z}} {\overline{w} } \).

Let \( z = a + bi, w = c + di \) then \[ \begin{aligned} \overline { z \times w} &= \overline { (ac-bd) + (ad+bc) i } \\ &= (ac-bd) - (ad+bc) i, \end{aligned}\] while \[ \begin{aligned} \overline{z} \times \overline{w} &= (a-bi) \times (c-di) \\ &= [( ac+ (-bi)(-di) ] + [a(-d) + (-b) c] i \\ &= (ac-bd) - (ad+bc) i. \end{aligned}\] Hence they are the same.

As for the division, let's use the language of norm and conjugates that we've learned. You can also do this using \( a, b, c, d \) as above.

\( \begin{aligned} \overline{ \left( \frac {z}{w} \right)} & = \overline{ \left( \frac {z \times \overline{w}} {w \times \overline{w}} \right) } & & \mbox{ multiplying by conjugates} \\ & = \overline{ \left(\frac {z \times \overline{w}} {N( w )} \right) }& &\mbox{definition of Norm} \\ & =\frac { \overline{ z \times \overline{w}} } { N(w )} & &\mbox{since Norm is a real value}\\ & = \frac {\overline{z} \times \overline{\bar{w}}}{N( w )}&& \mbox{since conjugation distributes over multiplication} \\ & = \frac {\overline{z} \times w }{N( w ) } && \mbox{property of conjugation} \\ & = \frac {\overline{z} \times w} { \overline{w} \times w}& &\mbox{ definition of Norm} \\ & = \frac {\overline{z}} {\overline{w} }& &\mbox{ cancellation} \\ \end{aligned} \)

You should easily verify that conjugation distributes over addition and subtraction.

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