This post is targeted at a Level 3 student.

After having learned about rational numbers and irrational numbers, which form the set of real numbers, the next extension that students see will be the complex numbers. This arises naturally when we’re looking to solve a quadratic equation. 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 $i^2 = -1$. If so, $i$ 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 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 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 $\frac {a+bi} {c+di} = \frac {(a+bi)(c-di)}{(c+di)(c-di)} = \frac {(ac-bd) + (-ad+bc)i}{c^2 + d^2 }$.

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 $b = 0$.

Solution: 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$\overline { z \times w} = \overline { (ac-bd) + (ad+bc) i } = (ac-bd) - (ad+bc) i$, while$\overline{z} \times \overline{w} = (a-bi) \times (c-di) = [( ac+ (-bi)(-di) ] + [a(-d) + (-b) c] i = (ac-bd) - (ad+bc) i.$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.

Test Yourself

1. Determine the 4 roots of the equation $x^4 + 1 = 0$, and find their real and imaginary parts.

2. Verify that the norm distributes over multiplication and division. Specifically, show that $N( z \times w ) = N( z ) \times N( w )$ and that $N\left( \frac {z}{w} \right) = \frac { N( z )} { N( w )}$. Give examples to show that the norm does NOT distribute over addition and subtraction.

3. If $\frac { (1+2i)(2+3i)}{8+i} = a + bi$, what is $a^2 +b^2$?
Hint: There is no need to determine the exact values of $a$ and $b$.

4. Determine the square root of $1-i$.

Note: We can show that the square root of $z = a + bi$ is equal to $\pm \left( \sqrt{ \frac {a + \sqrt{a^2+b^2}}{2}} + sgn(b) \sqrt{ \frac {-a + \sqrt{a^2 + b^2}}{2}} \right)$. Currently, our only way to show this is through brute force multiplication. We will be learning how to approach this problem otherwise.

1. Reblogged this on The Blog of Sigma(Sum).

2. nice and simple… but what is the practical meaning of a complex number??

• Can you clarify, what do you mean by practical?

Complex numbers are used as solutions to equations, for understanding fluid dynamics (and hence the all important air-conditioner), approaching relativistic effects (time is placed on the imaginary axis), etc.

Also, note that this is an introduction to the basic properties of complex numbers, and there have been no ‘applications’ mentioned.

If we cannot even imagine something like sqrt(-1) then why do and how do we use it?

• There are ways that we can picture it, like using the Polar form / Argand Diagram. There are people who even just use a line to represent it, though this line will clearly be different from the real number line that you are used to.

As you develop in mathematics, there are a lot of structures which may not have an obvious physical representation, but are still extremely useful. For example, it will be useful to be able to think in many n-dimensions, but we only have a good handle on 3 physical dimensions, or at most 4. It will become difficult to think in 6-dimensions, but it is still useful to know how to use it.

Does the following property exist?
| z1 – z2 | <= | z1 + z2 |

• This is an interesting one, which you should be able to figure it out.

Here’s a possible hint. Assume that the property exists, then if we replace z2 by -z2, we get that $| z1 + z2 \leq |z1 -z2|$, which hence tells us that the two absolute values must be the same.

hi everyone
1) ${x^4+1= (x^2)^2+1}$
so $x^2= i$
x= ${\sqrt{i}}$ or ${\sqrt{-i }}$ or -${\sqrt{i}}$ or -${\sqrt{-i}}$

2) N(z .w)=z .w x ${\overline{z .w}} = z . w$ .${\overline{w}}$. ${\overline{z}}= N(z) . N(w)$
same thing for z/w

3)${\frac{(1+2i)(2+3i)}{8+i}}$ =${\frac{(1+2i)(2+3i)}{8+i}}$ = ${\frac{-25+60i}{63}}$
a²+b²= ${\frac{985}{3969}}$

sorry I forgot how to use latex …

• To type latex on the blog (WordPress is quite fussy}, type it as $_latex code$, where the _ has been removed.

I did the following necessary edits to your latex
1. added a space after the work latex
2. added a $at the end of the code 3. Closed off some brackets { } that you had left open 4. Replaced some of your displayed squares by ^2. Latex doesn’t understand such images. Note: There is no need to place all of the code within { } . For example,$_latex a^2 + b^2 = c^2\$ would be sufficient.

• For 1, you cannot conclude that $x^2 = i$.
As an example, if we wanted to solve $y^2 = 1$, we cannot simply take square roots and say that $y=1$. This is an extremely common mistake when it comes to solving quadratic equations.

For 2, you are assuming that multiplication in the complex numbers commutes, i.e. that $z \times w = w \times z$. This has not been established as yet, though it would follow from the ‘definition’ of multiplication and the commutativity of the real numbers.

For 3, try and do it without calculating the exact values of a and b, as the hint says. I’m also not too clear on how you did the multiplication, and I believe that there are several calculation mistakes.

You are right Calvin.When we solve the equation $y^2-1=0$ we are left with either y+1=0 or y-1=0 . So we are left with +1 or -1.As we approach higher degree equations, it becomes even more necessary to sieve the real and complex solutions.