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# Complex Numbers

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:

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

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 conjugate of 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 norm of 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.

Note by Arron Kau
2 years, 7 months ago