**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.

## 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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