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Complex Numbers

A concept that will be very important later on is that of complex numbers. Complex numbers originally come from solutions of square roots, so we'll start there.

We all know how to solve equations like this $x^2 = 4$ . We don't even need to try particularly hard since we can just try values of $x$ until we find one that works. In this case, we know that both $2$ and $-2$ are solutions. Having two solutions is key here. If we have a positive solution to such an equation, we can also say its negative is a solution since a negative times a negative is a positive.

It seems then that when squaring a number we should always get a positive result. If we tried to solve the equation $x^2 = -1$ , we would find that there is no solution.

Or would we?

With our current knowledge, there is no known solution. But what if we were to just pretend one exists? Let's define a constant, $i$ , that is the "solution" to the equation $i^2 = -1$ . We can't express any meaningful value to $i$ , but we can still see how it behaves in equations.

One of the most basic and principle properties we will be able to study is the powers of $i$ :

i^0 = 1 \\[3ex] i^1=i \\[3ex] i^2=-1 \\[3ex] i^3=i\cdot i^2 = i \cdot (-1) = -i \\[3ex] i^4 = i^2 \cdot i^2 = (-1) \cdot (-1) = 1

From this we can see that:

i^0 = i^4 = i^8 = \cdots = i^{4n} = 1 \\[3ex] i^1 = i^5 = i^9 = \cdots = i^{4n + 1} = i \\[3ex] i^2 = i^6 = i^{10} = \cdots = i^{4n + 2} = -1 \\[3ex] i^3 = i^7 = i^{11} = \cdots = i^{4n + 3}= -i

It's quite suprising that the powers of $i$ follow this "circular" pattern. There's no number we know that behaves like this! Normally a number will simply grow larger and larger as we raise it to higher powers.

This "circular" idea will be our first step to a visual intuition of the complex numbers. We will imagine these numbers as lying on a circle that extends the standard number line:

A circular representation of the complex units on an Argan diagram

This starts to give us an intuition that complex numbers might be well visualised as 2-dimensional. Just as we would represent a point in 2 dimensions with an $x$ component and a $y$ component, we will represent our complex numbers as having a real component and an imaginary component.

We will write them as $a + bi$ where $a$ is the real component and $b$ is the imaginary component as it is the coefficient of $i$ .

Let's take a look at how we plot these numbers on the 2D plane:

A variety of example complex numbers plotted on an Argand diagram

So we can see how they plot just like 2D points.

In the same way we can calculate the distance of a point from the origin using pythagoras: $\sqrt{x^2 + y^2}$ , we can also calculate the distance of a complex number from $0$ as $\sqrt{a^2 + b^2}$ . This value we will call the modulus of the complex number and will be represented like this:

|a + bi| = \sqrt{a^2 + b^2}

Another concept we'll introduce is the complex conjugate. This will be represented with the $^*$ symbol and will simply make the complex part negative:

(a + bi)^* = a - bi

Here's an interesting property we'll show. For a complex number $z = a + bi$ we have:

z z^* = (a + bi)(a - bi) = a^2 - abi + abi - b^2i^2 = a^2 + b^2 = |z|^2

Polar Representation

We also know that we can represent points in 2D space using polar coordinates. We can do the same for complex numbers:

An example of a complex number in polar form, with an angle and a radius

We will represent the distance from the origin as $r$ and the angle from the postive-real line as $\theta$

Some knowledge of trigonometry shows us that for a complex number $z$ we can represent it in polar form as such:

z = r\cos{\theta} + ir\sin{\theta}

The polar system has another more interesting representation however, that is quite extraordinary!

We will begin in an unusual and seemingly unrelated place: the taylor expansion of $e^x$ :

e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots + \frac{x^n}{n!} + \ldots

We're going to check what happens when we introduce $i$ into the exponent:

e^{ix} = 1 + ix + \frac{(ix)^2}{2!} + \frac{(ix)^3}{3!} + \ldots + \frac{(ix)^n}{n!} + \ldots

Now we can apply our power rules of $i$ to simplify:

e^{ix} = 1 + ix - \frac{x^2}{2!} - i\frac{x^3}{3!} + \frac{x^4}{4!} + i\frac{x^5}{5!} - \frac{x^6}{6!} - i\frac{x^7}{7!} + \ldots

We've expanded the series out to more terms here to make the pattern more visible. Notice now we have some terms with an $i$ in them and some without. If we group the like terms and factor out the $i$ we get:

e^{ix} = \Bigl(1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \ldots \Bigr) + i\Bigl(x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \ldots \Bigr)

Our amazing discovery now comes as we realise that the two infinite series in the brackets are the taylor expansions for $\cos$ and $\sin$ respectively! So let's substitute them in:

e^{ix} = \cos{x} + i\sin{x}

That's quite the incredible formula! Trigonometric functions, exponentials and complex numbers all together!

Going back to our polar representation from earlier, we can now simplify it to:

z = r(\cos{\theta} + i\sin{\theta}) = r e^{i\theta}

This is a very useful representation that we will use heavily. Note that can also re-state our definition of modulus and conjugate in this form:

|re^{i\theta}| = r \\[3ex] (re^{i\theta})^* = re^{-i\theta}

Exercises

Exercise 1

Using the exponential form of the polar representation shown above, prove Euler's formula:

e^{i\pi} + 1 = 0

Show solution

Exercise 2

Prove the following properties about the conjugation operation:

(z^*)^* = z \\[3ex] (z + z^\prime)^* = z^* + z^{\prime*} \\[3ex] (z z^\prime)^* = z^* z^{\prime*}

Show solution

Exercise 3

Just as we showed some properties of the conjugate operator, now let's have a look at a property of the modulus.

Prove the following:

|z z^\prime| = |z| |z^\prime|

Show solution

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Additional Materials

About

Complex Numbers

Polar Representation

Exercises

Exercise 1

Exercise 2

Exercise 3