# High School Mathematics Extensions/Supplementary/Complex Numbers

## Introduction

Content Supplementary Chapters Basic Counting Polynomial Division Partial Fractions Summation Sign Complex Numbers Differentiation Problem Set Exercise Solutions Problem Set Solutions

Although the real numbers can, in some sense, represent any natural quantity, they are in another sense incomplete. We can write certain types of equations with real number coefficients which we desire to solve, but which have no real number solutions. The simplest example of this is the equation:

${\displaystyle {\begin{matrix}x^{2}+1&=&0\\x^{2}&=&-1\\x&=&{\sqrt {-1}}\end{matrix}}}$

Your high school math teacher may have told you that there is no solution to the above equation. He/she may have even emphasised that there is no real solution. But we can, in fact, extend our system of numbers to include the complex numbers by declaring the solution to that equation to exist, and giving it a name: the imaginary unit, ${\displaystyle i}$.

Let's imagine for this chapter that ${\displaystyle i={\sqrt {-1}}}$ exists. Hence x = i is a solution to the above question, and ${\displaystyle i^{2}=-1}$.

A valid question that one may ask is "Why?". Why is it important that we be able to solve these quadratics with this seemingly artificial construction? It is interesting delve a little further into the reason why this imaginary number was introduced in the first place - it turns out that there was a valid reason why mathematicians realized that such a construct was useful, and could provide deeper insight.

The answer to the question lies not in the solution of quadratics, but rather in the solution of the intersection of a cubic and a line. The mathematician Cardano managed to come up with an ingenious method of solving cubics - much like the quadratic formula, there is also a formula that gives us the roots of cubic equations, although it is far more complicated. Essentially, we can express the solution of a cubic ${\displaystyle x^{3}=3px+2q}$ in the form

${\displaystyle x={\sqrt[{3}]{q+{\sqrt {q^{2}-p^{3}}}}}+{\sqrt[{3}]{q-{\sqrt {q^{2}-p^{3}}}}}}$

An unsightly expression, indeed!

You should be able to convince yourself that the line ${\displaystyle y=3px+2q}$ must always hit the cubic ${\displaystyle y=x^{3}}$. But try solving some equation where ${\displaystyle q^{2}, and you run into a problem - the problem is that we are forced to deal with the square root of a negative number. But, we know that in fact there is a solution for x; for example, ${\displaystyle x^{3}=15x+4}$ has the solution x = 4.

It became apparent to the mathematician Bombelli that there was some piece of the puzzle that was missing - something that explained how this seemingly perverse operation of taking a square root of a negative number would somehow simplify to a simple answer like 4. This was in fact the motivation for considering imaginary numbers, and opened up a fascinating area of mathematics.

The topic of Complex numbers is very much concerned with this number i. Since this number doesn't exist in this real world, and only lives in our imagination, we call it the imaginary unit. (Note that ${\displaystyle i}$ is not typically chosen as a variable name for this reason.)

### The imaginary unit

As mentioned above

${\displaystyle {\begin{matrix}i^{2}&=&-1\end{matrix}}}$.

Let's compute a few more powers of i:

${\displaystyle {\begin{matrix}i^{1}&=&i\\i^{2}&=&-1\\i^{3}&=&-i\\i^{4}&=&1\\i^{5}&=&i\\i^{6}&=&-1\\&{\mbox{...}}&\end{matrix}}}$

As you may see, there is a pattern to be found in this.

### Exercises

1. Compute ${\displaystyle i^{25}}$
2. Compute ${\displaystyle i^{100}}$
3. Compute ${\displaystyle i^{1000}}$
Exercise Solutions

### Complex numbers as solutions to quadratic equations

${\displaystyle {\begin{matrix}x^{2}-6x+13&=&0\\x&=&{\frac {6\pm {\sqrt {36-4\times 13}}}{2}}\\x&=&{\frac {6\pm {\sqrt {-16}}}{2}}\\x&=&{\frac {6\pm {\sqrt {-1}}{\sqrt {16}}}{2}}\\x&=&{\frac {6\pm 4i}{2}}\\x&=&3+2i\ ,\ 3-2i\\\end{matrix}}}$

The x we get as a solution is what we call a complex number. (To be nitpicky, the solution set of this equation actually has two complex numbers in it; either is a valid value for x.) It consists of two parts: a real part of 3 and an imaginary part of ${\displaystyle \pm 2}$. Let's call the real part a and the imaginary part b; then the sum ${\displaystyle a+bi=3\pm 2i}$ is a complex number.

Notice that by merely defining the square root of negative one, we have already given ourselves the ability to assign a value to a much more complicated, and previously unsolvable, quadratic equation. It turns out that 'any' polynomial equation of degree ${\displaystyle n}$ has exactly ${\displaystyle n}$ zeroes if we allow complex numbers; this is called the Fundamental Theorem of Algebra.

We denote the real part by Re. E.g.:

${\displaystyle \mathrm {Re} (x)=3}$

and the imaginary part by Im. E.g.:

${\displaystyle \mathrm {Im} (x)=\pm 2}$

Let's check to see whether ${\displaystyle x=3+2i}$ really is solution to the equation:

${\displaystyle {\begin{matrix}x&=&3+2i&\\x^{2}&=&(3)^{2}+2(3)(2i)+(2i)^{2}\\&=&5+12i\\x^{2}-6x+13&=&5+12i-6(3+2i)+13\\&=&0\\\end{matrix}}}$

### Exercises

1. Convince yourself that x = 3 - 2i is also a solution to the equation.
2. Plot the points A(3, 2) and B(3, -2) on a XY plane. Draw a line for each point joining them to the origin.
3. Compute the length of AO (the distance from point A to the Origin) and BO. Denote them by r1 and r2 respectively. What do you observe?
4. Compute the angle between each line and the x-axis and denote them by ${\displaystyle \phi _{1}}$ and ${\displaystyle \phi _{2}}$. What do you observe?
5. Consider the complex numbers:
${\displaystyle {\begin{matrix}z&=&r_{1}\cdot (\cos {\phi _{1}}+i\sin {\phi _{1}})\\w&=&r_{2}\cdot (\cos {\phi _{2}}+i\sin {\phi _{2}})\\\end{matrix}}}$

Substitute z and w into the quadratic equation above using the values you have computed in Exercise 3 and 4. What do you observe? What conclusion can you draw from this?

1. Find the complex solutions to the equation ${\displaystyle x^{2}-2x+2=0}$. Perform the same steps as above. What can you conclude here?

## Arithmetic with complex numbers

Adding and multiplying two complex number together turns out to be quite straightforward. Let's illustrate with a few examples. Let x = 3 - 2i and y = 7 + 11i, and we do addition first

 ${\displaystyle x+y\!}$ ${\displaystyle =\!}$ ${\displaystyle (3+7)+(-2+11)i\!}$ ${\displaystyle =\!}$ ${\displaystyle 10+9i\!}$

and now multiplication

 ${\displaystyle x\cdot y\!}$ ${\displaystyle =\!}$ ${\displaystyle (3-2i)\cdot (7+11i)\!}$ ${\displaystyle =\!}$ ${\displaystyle 3\cdot 7+3\cdot 11i-2i\cdot 7-2\cdot 11\cdot i^{2}\!}$ ${\displaystyle =\!}$ ${\displaystyle 43+19i\!}$

Let's summarise the results here.

• When adding complex numbers we add the real parts with real parts, and add the imaginary parts with imaginary parts.
• When multiplying two complex numbers together, we use normal expansion. Whenever we see i2 we put in its place -1. We then collect like terms.

But how do we calculate:

${\displaystyle {\frac {3+2i}{7-{\sqrt {5}}i}}}$

Note that the square root is only above the 5 and not the i. This is a little bit tricky, and we shall cover it in the next section.

### Exercises:

${\displaystyle {\begin{matrix}x&=&3-2i\\y&=&3+2i\end{matrix}}}$

Compute:

1. x + y
2. x - y
3. x2
4. y2
5. xy
6. (x + y)(x - y)

### Division

One way to calculate:

${\displaystyle {\frac {1}{2{\sqrt {3}}+{\sqrt {2}}}}}$

is to rationalise the denominator:

${\displaystyle {\frac {1}{2{\sqrt {3}}+{\sqrt {2}}}}={\frac {2{\sqrt {3}}-{\sqrt {2}}}{(2{\sqrt {3}}+{\sqrt {2}})(2{\sqrt {3}}-{\sqrt {2}})}}={\frac {2{\sqrt {3}}-{\sqrt {2}}}{10}}}$

Utilising a similar idea, to calculate

${\displaystyle {\frac {3+2i}{7-{\sqrt {5}}i}}}$

we realise the denominator.

${\displaystyle z\ =\ {\frac {3+2i}{7-{\sqrt {5}}i}}}$
${\displaystyle z\ =\ {\frac {3+2i}{7-{\sqrt {5}}i}}\times {\frac {7+{\sqrt {5}}i}{7+{\sqrt {5}}i}}}$

The denominator is the sum of two squares. We get:

${\displaystyle z\ =\ {\frac {(3+2i)\times (7+{\sqrt {5}}i)}{49+5}}}$
${\displaystyle z\ =\ {\frac {21-2{\sqrt {5}}}{54}}+{\frac {14+3{\sqrt {5}}}{54}}i}$

If somehow we can always find a complex number whose product with the denominator is a real number, then it's easy to do divisions.

If

${\displaystyle z\ =\ a+ib}$

and

${\displaystyle w\ =\ a-ib}$

Then zw is a real number. This is true for any 'a' and 'b' (provided they are real numbers).

### Exercises

Convince yourself that the product of zw is always a real number.

### Complex Conjugate

The exercise above leads to the idea of a complex conjugate. The complex conjugate of a + ib is a - ib. For example, the conjugate of 2 + 3i is 2 - 3i. It is a simple fact that the product of a complex number and its conjugate is always a real number. If z is a complex number then its conjugate is denoted by ${\displaystyle {\bar {z}}}$. Symbolically if

z = a + ib

then,

${\displaystyle {\bar {z}}=a-ib}$

The conjugate of 3 - 9i is 3 + 9i.

The conjugate of 100 is 100.

The conjugate of 9i - 20 is -20 - 9i.

Conjugate laws

Here are a few simple rules regarding the complex conjugate

${\displaystyle {\overline {z+w}}={\bar {z}}+{\bar {w}}}$

and

${\displaystyle {\overline {zw}}={\bar {z}}{\bar {w}}}$

The above laws simply says that the sum of conjugates equals the conjugate of the sum; and similarly, the conjugate of the product equals the product of the conjugates.

Consider this example:

${\displaystyle (3+2i)+(89-100i)=92-98i}$

and we can see that

${\displaystyle {\overline {92-98i}}=92+98i}$

which equals to

${\displaystyle {\overline {3+2i}}+{\overline {89-100i}}=3-2i+89+100i=92+98i}$

This confirms the addition conjugate law.

### Exercise

Convince yourself that the multiplication law is also true.

## The complex root

Now that you are equipped with all the basics of complex numbers, you can tackle the more advanced topic of root finding.

Consider the question:

${\displaystyle {\begin{matrix}z&=&-3+4i\\w&=&{\sqrt {z}}\end{matrix}}}$

Express w in the form of a + ib.

That is easy enough.

${\displaystyle {\begin{matrix}w&=&{\sqrt {z}}\\w^{2}&=&z\\w&=&a+ib\\w^{2}&=&a^{2}-b^{2}+2abi\\\\-3&=&a^{2}-b^{2}&{\mbox{(1)}}&\\4&=&2ab&{\mbox{(2)}}&\\\end{matrix}}}$

Solve (1) and (2) simultaneously to work out a and b.

Observe that if, after solving for a and b, we replace them with -a and -b respectively, then they would still satisfy the two simultaneous equations above, we can see that (as expected) if w = a + ib satisfies the equation w2 = z, then so will w = -(a + ib). With real numbers, we always take the non-negative answer and call the solution ${\displaystyle {\sqrt {x}}}$. However, since there is no notion of "greater than" or "less than" with complex numbers, there is no such choice of ${\displaystyle {\sqrt {z}}}$. In fact, which square root to take as "the" value of ${\displaystyle {\sqrt {z}}}$ depends on the circumstances, and this choice is very important to some calculations.

### info -- Finding the square root

Finding the root of a real number is a very difficult problem to start with. Most people have no hope of finding a close estimate of ${\displaystyle {\sqrt {2}}}$ without the help of a calculator. The modern method of approximating roots involves an easy to understand and ingenius piece of mathematics called the Taylor series expansion. The topic is usually covered in first year university maths as it requires an elementary understanding of an important branch of mathematics called calculus.

The Newton-Raphson method of root finding is also used extensively for this purpose.

Now consider the problem

${\displaystyle {\begin{matrix}z&=&-2+2i\\w&=&z^{1/3}\end{matrix}}}$

Express w in the form of "a + ib".

Using the methodology developed above we proceed as follows,

${\displaystyle {\begin{matrix}w&=&z^{1/3}\\w^{3}&=&z\\\\w&=&(a+ib)\\w^{3}&=&(a^{2}-b^{2}+2abi)\times (a+ib)\\z&=&(a^{3}-3ab^{2})+i(3a^{2}b-b^{3})\\\\-2&=&a^{3}-3ab^{2}\qquad (1)\\2&=&3a^{2}b-b^{3}\qquad (2)\\\end{matrix}}}$

It turns out that the simultaneous equations (1) & (2) are hard to solve. Actually, there is an easy way to calculate the roots of complex numbers called the De Moivre's theorem, it allows us to calculate the nth root of any complex number with ease. But to set the method, we need understand the geometric meaning of a complex number and learn a new way to represent a complex number.

### Exercises

1. Find (3 + 3i)1/2
2. Find (1 + 1i)1/2
3. Find i1/3

## The complex plane

### Complex numbers as ordered pairs

It is worth noting, at this point, that every complex number, a + bi, can be completely specified with exactly two real numbers: the real part a, and the imaginary part b. This is true of every complex number; for example, the number 5 has real part 5 and imaginary part 0, while the number 7i has real part 0 and imaginary part 7. We can take advantage of this to adopt an alternative scheme for writing complex numbers: we can write them as ordered pairs, in the form (a, b) instead of a+bi.

${\displaystyle {\begin{matrix}{\mbox{Instead of}}&{\mbox{We could write}}\\5+4i&(5,4)\\3i&(0,3)\\{\frac {4+5i}{3}}&({\frac {4}{3}},{\frac {5}{3}})\\42&(42,0)\\{\sqrt {2}}+{\sqrt {2}}i&({\sqrt {2}},{\sqrt {2}})\\\end{matrix}}}$

These should look familiar: they are exactly like the ordered pairs we use to represent points in the plane. In fact, we can use them that way; the plane which results is called the complex plane. We refer to its x axis as the real axis, and to its y axis as the imaginary axis.

### The complex plane

We can see from the above that a single complex number is a point in the complex plane. We can also represent sets of complex numbers; these will form regions on the plane. For example, the set

${\displaystyle \{a+bi|-1\leq a\leq 1,-1\leq b\leq 1\}}$

is a square of edge length 2 centered at the origin (See following diagram).

### Complex-valued functions

Just as we can make functions which take real values and output real values, so we can create functions from complex numbers to real numbers, or from complex numbers to complex numbers. These latter functions are often referred to as complex-valued functions, because they evaluate to (output) a complex number; it is implicit that their argument (input) is complex as well.

Since complex-valued functions map complex numbers to other complex numbers, and we have already seen that complex numbers correspond to points on the complex plane, we can see that a complex-valued function can turn regions on the complex plane into other regions. A simple example: the function

${\displaystyle f(z)=z+(0+1i)}$

takes a point in the complex plane and shifts it up by 1. If we apply it to the set of points making up the square above, it will move the entire square up one, so that it "rests" on the x-axis.

{To make more complicated examples, I will first have to go back and introduce the polar representation of complex numbers. Makes for much more interesting functions, :-) You can use the diagrams below or modify them to make new diagrams. I will make links to these diagrams in other places in Wikibooks:math. In the 2nd diagram showing the point, r=4 and theta= 50 degrees. These types of diagrams can be used to introduce phasors, which are notations for complex numbers used in electrical engineering.}

## de Moivre's Theorem

${\displaystyle (e^{i\theta })^{n}=(\cos \theta +j\sin \theta )^{n}=\cos n\theta +j\sin n\theta }$

For example, say we have ${\displaystyle z=1+j}$. We may now write this in polar form,

${\displaystyle z=|z|e^{j\arg z}={\sqrt {2}}e^{{\frac {\pi }{4}}j}={\sqrt {2}}\left(\cos \left({\frac {\pi }{4}}\right)+i\sin \left({\frac {\pi }{4}}\right)\right)}$

Using de Moivre's theorem, we can deduce,

${\displaystyle z^{2}=2\left(\cos \left({\frac {\pi }{2}}\right)+j\sin \left({\frac {\pi }{2}}\right)\right)=2j}$
${\displaystyle z^{3}=2{\sqrt {2}}\left(\cos \left({\frac {3\pi }{4}}\right)+j\sin \left({\frac {3\pi }{4}}\right)\right)={\sqrt {2}}\left({\frac {\sqrt {2}}{2}}j-{\frac {\sqrt {2}}{2}}\right)=2j-2}$
${\displaystyle z^{4}=4\cdot -1=-4}$

etc.

## Complex root of unity

The complex roots of unity to the nth degree is the set of solutions to the equation ${\displaystyle x^{n}=1}$. Clearly they all have magnitude 1. They form a cyclic group under multiplication. For any given ${\displaystyle n}$, there are exactly ${\displaystyle n}$ many of them, and they form a regular n-gon in the complex plane over the unit circle.

A closed form solution can be given for them, by use of Euler's formula:

${\displaystyle u^{n}=\cos(2\pi \cdot j/n)+i\sin(2\pi \cdot j/n)\quad 0\leq j

The sum of the ${\displaystyle n}$th roots of unity is equal to 0, except for ${\displaystyle n=1}$, where it is equal to 1.

The product of the ${\displaystyle n}$th roots of unity alternates between -1 and 1.

## Problem set

Simplify ${\displaystyle (1-i)^{2i}}$

First, write ${\displaystyle (1-i)}$, as

${\displaystyle |1-i|e^{i\arg(1-i)}={\sqrt {2}}e^{-i{\frac {\pi }{2}}}}$

Then,

${\displaystyle (1-i)^{2i}=\left({\sqrt {2}}e^{-i{\frac {\pi }{2}}}\right)^{2i}=2^{i}e^{\frac {\pi }{2}}}$