Chapter 1 - Complex Numbers

Section 1: Basic Definitions

Complex numbers have their origin in the theory of algebraic equations (or polynomials). It became apparent that in plenty of cases, no solution existed among the usual set of numbers (positive numbers, natural numbers, integers, rational numbers, real numbers). A simple example is below:

$x^2 + 1 = 0$

Indeed, any possible $x^2$ as we know it would be a positive number or zero. Since here we want $x^2$ to be -1, we are stuck!

One option would be to just give up on finding a solution! But failure is not an option at least for mathematicians. Before discussing the new system that mathematicians came up with, let's remind ourselves of the existing set of numbers,

$\bullet$ Positive numbers: $\mathbb{P} = \{1, 2, 3, ...\}$

$\bullet$ Natural numbers: $\mathbb{N} = \{0, 1, 2, 3, ...\}$

$\bullet$ Integers: $\mathbb{Z} = \{..., -3, -2, -2, 0, 1, 2, 3, ...\}$

$\bullet$ Rational numbers: $\mathbb{Q} = \{\frac{m}{n}|m\in\mathbb{Z},n\in\mathbb{P}\}$

$\bullet$ Real numbers: $\mathbb{R} = \mathbb{Q}\bigcup\{...,\sqrt{2},...,e,...,\pi,...,\frac{e}{\pi}\}$

None of these familiar systems have a solution for $x^2=-1$. In a desparate attempt to find a solution, we define a new number, denoted by $j$ (usually $i$) such that,

$j^2 = -1$ or $j=\sqrt{-1}$

We know that such a number does not exist amongst real numbers and hence we call it an imaginary number. Aside from the strange behavior that square of $j$ is -1, it behaves like any regular number. Let's do some exercises:

Example 1.1

What is the value of $j^3$? We could do

$j^3 = j\times j\times j = (j^2)\times j = -1\times j = -j$

Or we could do it in code as shown below

Exercise 1.1 (30 points)

Simplify these analytically and then verify with code
a. $j^0$, $j^4$, $j^5$, $j^6$, $j^7$ and $j^8$
b. $j^{39}$, $j^{174}$, $j^{252}$ and $j^{317}$

$j^0$ = 1
$j^4$ == $\sqrt{-1} \times \sqrt{-1} \times \sqrt{-1} \times \sqrt{-1}$ == $-1 \times -1$ == 1
$j^5$ == $\sqrt{-1} \times j^4$ == $1 \times \sqrt{-1}$ == $\sqrt{-1}$
$j^6$ == $j^5 \times \sqrt{-1}$ == $\sqrt{-1} \times \sqrt{-1}$ == -1
$j^7$ == $j^6 \times \sqrt{-1}$ == $-1 \times \sqrt{-1}$ == -$\sqrt{-1}$
$j^8$ == $j^7 \times \sqrt{-1}$ == $-\sqrt{-1} \times \sqrt{-1}$ == $-(-1)$ == 1
$j^{39}$ == $(j^4)^9$ * $j^3$
$j^{174}$
$j^{252}$
$j^{317}$

$\textbf{Points:}$ 2 points per simplification done analytically and additional 1 point for verifying it with code.

We can also multiply $j$ with another number, for example $2\times j$. These numbers are called $\textbf{imaginary numbers}$.

We can, in fact, add a real number and an imaginary number, for instance $3+5\times j$, and you get a number that is neither real nor imaginary. Such a "hybrid" number is called a $\textbf{complex number}$.

Definition 1: Complex Number

A complex number is an expression $c = a + b\times j = a+bj$, where a, b are two real numbers; a is called the real part of c, whereas b is its imaginary part. The set of all complex numbers will be denoted by $\mathbb{C}$. When the $\times$ is understood, it is omitted from writing.

Example 1.2

Let $c_1 = 3 - j$ and $c_2 = 1 + 4j$. We want to compute $c_1 + c_2$ and $c_1 \times c_2$

Exercise 1.2 (20 points)

Now instead of using builtin functions to add and multiply complex numbers, write your own functions. The functions must take as inputs two complex numbers.

Exercise 1.3 (16 points)

Calculate $c_1+c_2$ and $c_1\times c_2$ using the functions you just created. Verify your results against the ones you get from builtin functions.
a. $c_1 = -3+j$ and $c_2 = 2-4j$
b. $c_1 = -4+7j$ and $c_2 = 5-10j$
c. $c_1 = 4+12j$ and $c_2 = -(3-15j)$
d. $c_1 = 5j$ and $c_2 = -(-9+i)$
e. $c_1 = 7j$ and $c_2 = -5+2j$
f. $c_1 = 1-5j$ and $c_2 = -9+2j$
g. $c_1 = 4+j$ and $c_2 = 2+3j$
h. $c_1 = 1-8j$ and $c_2 = 1+8j$

$\textbf{Points:}$ 2 points each for computation and verification

Fundamental Theorem of Algebra

Every polynomial equation of one variable with complex coefficients has a complex solution

Exercise 1.4 (9 points)

Verify the following:
a. $(-1+j)$ is a solution to the polynomial equation $x^2+2x+2=0$.
b. $(-1-3j)$ is a solution to the polynomial equation $x^2+2x+10=0$.
c. $(2-\sqrt{3}j)$ is a solution to polynomial equation $x^2-4x+7=0$.

Section 2. The Algebra of Complex Numbers

To understand complex numbers better we will soon discuss their geometrical interpretation. However, before we can do that, we'll write complex numbers as a tuple of values made up of two real numbers.

We know that $a+bj$ is made up of a real part $a$ and an imaginary part $b$. We can therefore redefine the complex number as an ordered pair of reals: $c \mapsto (a, b)$

Therefore, ordinary real numbers can be identified with pairs $(a, 0)$. As a result, $a \mapsto (a,0)$

whereas imaginary numbers will be pairs $(0, b)$. In particular, $bj \mapsto (0,b)$

$\textbf{Addition}$ is rather obvious: it adds pairs componentwise: $(a_1 + b_1) + (a_2 + b_2) = (a_1 + a_2, b_1 + b_2)$

$\textbf{Multiplication}$ is little trickier: $(a_1, b_1)\times(a_2, b_2) = (a_1, b_1)(a_2, b_2) = (a_1a_2-b_1b_2, a_1b_2+a_2b_1)$

Let's see if the multiplication formula above works: $j\times j = (0, 1)\times(0,1) = (0-1, 0+0) = (-1,0) = -1$

Using addition and multiplication, we can write any complex number in the usual form: $c = (a,b) = (a,0)+(0,b) = (a,0)+(b,0)\times(0,1)=a+bj$

In summary, we have traded one oddity for another: $j$ was previously quite mysterious, whereas now it is just $(0,1)$ and we can essentially forget about $j$.

Therefore, a complex number is nothing more than just an ordered pair of ordinary real numbers!

Example 2.1

Let $c_1=(3,-2)$ and $c_2=(1,2)$. Let us multiply them using the aforementioned rule:
$c_1 \times c_2 = (3\times1-(-2)\times2, 3\times2+(-2)\times1) = (3+4, 6+(-2)) = (7,4) = 7+4j$

Exercise 2.1 (15 + 4 points)

Write a function to multiply two complex numbers using the tuple format and give the result in a tuple format. Compute the product of,
a. $c_1=(-3,-1)$ and $c_2=(1,-2)$
b. $c_1=(5,3)$ and $c_2=(2,-7)$
c. $c_1=(4,-3)$ and $c_2=(4,-3)$
d. $c_1=(5,3)$ and $c_2=(2,-4)$

$\textbf{Points:}$ 15 points for the program and 1 point each for the computation

So far we have a set of numbers and two operations: addition and multiplication. Both operations are $\textbf{commutative}$, meaning that for arbitrary complex numbers $c_1$ and $c_2$,
$c_1 + c_2 = c_2 + c_1$
and
$c_1 \times c_2 = c_2 \times c_1$

Both operations are also $\textbf{associative}$:
$(c_1 + c_2) + c_3 = c_1 + (c_2 + c_3)$
and
$(c_1\times c_2)\times c_3 = c_1\times(c_2\times c_3)$

Exercise 2.2 (15 + 3 points)

Verify that multiplication of complex numbers is associative for the following:
a. $c_1=(-3,-1)$, $c_2=(1,-2)$ and $c_3=(5,3)$
b. $c_1=(2,-7)$, $c_2=(4,-3)$ and $c_3=(4,-3)$
c. $c_1=(5,3)$, $c_2=(2,-4)$ and $c_3=(-3,1)$

$\textbf{Points:}$ 15 points for the program and 1 point each for the computation

Lastly, multiplication $\textbf{distributes}$ over addition: for all $c_1, c_2, c_3$, we have,
$c_1\times(c_2+c_3) = (c_1\times c_2) + (c_1\times c_3)$

Exercise 2.3 (15 + 3 points)

Verify the property that multiplication distributes over addition for the following,
a. $c_1=(-3,-1)$, $c_2=(1,-2)$ and $c_1=(5,3)$
b. $c_1=(2,-7)$, $c_2=(4,-3)$ and $c_3=(4,-3)$
c. $c_1=(5,3)$, $c_2=(2,-4)$ and $c_3=(-3,1)$

$\textbf{Points:}$ 15 points for the program and 1 point each for the computation

Having discussed addition and multiplication, we now discuss their complementary operations: subtraction and division

$\textbf{Subtraction}$ is straight forward and defined componentwise:
$c_1 - c_2 = (a_1, b_1) - (a_2, b_2) = (a_1 - a_2, b_1 - b_2)$

$\textbf{Division}$ is somewhat involved. We want to compute,
$(x,y) = \frac{(a_1, b_1)}{(a_2, b_2)}$

After some simple substitutions and computations we get,
$x = \frac{a_1a_2 + b_1b_2}{a_2^2 + b_2^2}$ and
$y = \frac{a_2b_1 - a_1b_2}{a_2^2 + b_2^2}$

Note that both $x$ and $y$ are calculated using the same denominator, namely, $a_2^2 + b_2^2$. We will discuss this quantity soon.

Example 2.2

Let $c_1=-2+j$ and $c_2=1+2j$. We will compute $\frac{c_1}{c_2}$. In this case, $a_1=-2$, $b_1=1$, $a_2=1$, and $b_2=2$. Therefore,
$a_2^2 + b_2^2 = 1^2 + 2^5 = 5$
$a_1a_2 + b_1b_2 = -2\times1 + 1\times2 = 0$
$a_2b_1 - a_1b_2 = 1\times1 - (-2)\times2 = 1+4 = 5$

The answer then is $(\frac{0}{5}, \frac{5}{5}) = (0, 1) = j$

Exercise 2.4 (15 + 3 points)

Write a function to calculate $\frac{c_1}{c_2}$,
a. $c_1 = 3j$ and $c_2 = -1-j$
b. $c_1 = 2-7j$ and $c_2 = 5+3j$
c. $c_1 = 5+3j$ and $c_2 = 2-4j$

$\textbf{Points:}$ 15 points for the program and 1 point each for the computation

Let us go back to discuss the denominator of our division operation above.

Real numbers have a unary operation called the $\textbf{absolute value}$ given by,
$|a| = +\sqrt{a^2}$

This operation is generalized in the complex domain as follows,
$|c| = |a+bj| = +\sqrt{a^2+b^2}$

This quantity is known as the $\textbf{modulus}$ of a complex number.

Example 2.3

What is the modulus of $c = 1-j$?
$|c| = |1-j| = +\sqrt{1^2 + (-1)^2} = \sqrt{2}$

Exercise 2.5 (10 + 5 points)

Write a function to compute the modulus of the following,
a. $5+3j$
b. $2-4j$
c. $6+4j$
d. $7-5j$
e. $4-3j$

$\textbf{Points:}$ 10 points for the program and 1 point each for the computation