LATMATH: Complex Numbers (2024)

In our world, we mostly observe real numbers. Through geometry and algebra, we find uses for numbers to explain many things. However, our understanding of numbers currently come with a cost. We have been able to solve equations like $3x + 1 = 4$ but we've come to believe that we can't solve equations like $x^2 = -1.$ Some might question the point of solving such an equation. On the surface level, if there is no solution, why would we bother extending cases like these? Questions like these were the reason why complex numbers were defined and created. We will observe the many uses as time goes on, but for now we should start by going over the main features of these numbers.

Complex Numbers

Up to this point, we've spoken about real numbers. Every number you've encountered up until this point has been a real number. $\pi,$ $e,$ $\frac{3}{5},$ $\sqrt{2},$ $9$ and so on. We've probably accepted that some equations just can't be solved. For example, solving this equation: $x^2 + x + 1 = 0.$ Let's try using our quadratic formula, which gives us $x = \frac{-1 \pm \sqrt{-3}}{2}.$ Well, it appears that this is as far as we can go. This equation doesn't have any real solutions since we cannot take the square root of a negative number.

Questions like these are the reasons we have created complex numbers. Every time we deal with an equation like this, we may result in taking the square root of a negative number. This all boils down to being able to find a solution to $x^2 = -1.$ Enter the world of complex numbers. Numbers that may not necessarily be observed in our real world. Our equation $x^2 = -1$ will clearly not have a real solution, so we need to imagine one. So, we call this number the imaginary unit, denoted as $i.$ $i$ has the property that $i^2 = -1.$

Now that we've define it, we can extend this to solve general square roots. For example, $\sqrt{-9}$ can be expressed as $\sqrt{9} \sqrt{-1} = 3i.$ Great, we're now able to find the square roots of negative numbers. But this brings up an important question, why can't we just say something like $\sqrt{9} = \sqrt{(-3)(-3)} = \sqrt{-3} \sqrt{-3} = 3i^2 = -3.$ So what's the issue with treating this unit with our previously defined understanding of square roots?

It's hard to grasp why this statement is wrong, and depending on how you define this operation, you may find this difficult to prove wrong. Indeed, the problem lies with the definition the square root of distributing over multiple negative numbers. We cannot completely define $\sqrt{(-1)(-1)}$ as $\sqrt{-1}\sqrt{-1}$ as this property cannot hold, however we can still observe that the result is simply a different square root of 1.

We've digressed a little from the main point, however. From what we understand, let's now approach solving the equation $x^2 + x + 1 = 0.$ We can first complete the square to get us $\left(x + \frac{1}{2} \right)^2 + \frac{3}{4} = 0 \Rightarrow \left(x + \frac{1}{2} \right) = -\frac{3}{4}.$ We are now faced with the same problem we had before, that is finding the square roots of this negative number. So, we get $x + \frac{1}{2} = \pm \sqrt{- \frac{3}{4}},$ to which we can then simplify as $x = -\frac{1}{2} + i\frac{\sqrt{3}}{2}.$ But now we've encountered a completely different problem. What does it mean to add an imaginary number to a real number.

Well, to understand what exactly this means, it's best to think about it at the present moment as a different like terms. We'll cover the arithmetic of complex numbers momentarily, however what we currently have will be enough to consider an example of solving another quadratic equation.

The previous methods of completing the square or using the quadratic formula will work perfectly fine. The only difference is now we may define solutions when the discriminant is negative.

Example: Solve for $x$ if $x^2 - 2x + 10 = 0.$

Let's first consider the method of using the quadratic formula, that is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{2 \pm \sqrt{-36}}{2}.$ Since we have $\sqrt{-36} = 6i,$ we can continue as $\frac{2 \pm 6i}{2} = 1 \pm 3i.$

Separately, we can consider the method of completing the square.

$ \begin{align} x^2 -2x + 10 &= 0\\ \,\\ x^2 - 2x + 1 + 9 &= 0\\ \,\\ (x - 1)^2 + 9 &= 0\\ \,\\ (x-1)^2 &= -9\\ \,\\ x-1 &= \pm \sqrt{-9}\\ \,\\ x - 1 &= \pm 3i\\ \,\\ x &= 1 \pm 3i. \end{align} $

Now, as promised, let's consider what it means to add a real number to an imaginary number. As I mentioned earlier, we can think of these as like terms. In fact, we can summarise the general working for complex arithmetic as the same as regular arithmetic except in this case any time we have a non-unit power of $i,$ we must consider rewriting. For the time being, we will consider integer powers of $i,$ mostly around $i^2 = -1.$

We can generalise a complex number $z$ as having a real and imaginary component. That is to say $z = x + iy$ where $x$ and $y$ are both real numbers (not complex themselves). With this general case, let's consider the following properties.

Consider two complex numbers $z_1 = x_1 + iy_1$ and $z_2 = x_2 + iy_2.$ We then have

$z_1 + z_2 = (x_1 + iy_2) + (x_2 + iy_2) = (x_1 + x_2) + i(y_1 + y_2);$

$z_1 \times z_2 = (x_1 + iy_1)(x_2 + iy_2) = (x_1x_2 - y_1y_2) + i(x_1y_2 + x_2y_1).$

Notably, the first definition follows from basic addition and factorisation. The second one follows from the expansion of a binomial product, except this time we have $i^2 = -1.$ You can try yourself and when you encounter $i^2$ make sure to replace that with $-1.$ Furthermore, we separate the real and imaginary parts to make it clear. Remember, $x_1, x_2, y_1$ and $y_2$ are all real numbers, so any product or sum/difference of them will remain real. If this is confusing, make sure to read through the following proof of these properties carefully.

Consider two complex numbers $z_1 = x_1 + iy_1$ and $z_2 = x_2 + iy_2.$

For this first property, it's just simple algebraic manipulation.

$ \begin{align} z_1 + z_2 &= x_1 + iy_1 + x_2 + iy_2\\ \,\\ &= x_1 + x_2 + iy_1 + iy_2\\ \,\\ &= (x_1 + x_2) + i(y_1 + y_2). \end{align} $
See Also
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We will do this next one by considering binomial expansion and the fact that $i^2 = -1.$

$ \begin{align} z_1 \times z_2 &= (x_1 + iy_1)(x_2 + iy_2)\\ \,\\ &= x_1 x_2 + ix_1y_2 + ix_2y_1 + i^2y_1y_2\\ \,\\ &= x_1 x_2 + ix_1y_2 + ix_2y_1 - y_1y_2\\ \,\\ &= (x_1 x_2 - y_1y_2) + i(x_1y_2 + x_2y_1). \end{align} $

As an interesting way to evaluate large powers of $i,$ it may be useful to notice that the powers cycle along $i = i,$ $i^2 = -1,$ $i^3 = -i,$ $i^4 = 1.$ This pattern repeats $i^5 = i,$ $i^6 = -1,$ $i^7 = -i$ and so on.

Further Arithmetic of Complex Numbers

We've covered the basics of addition and multiplication, but there's more to consider when working with complex numbers. Generally, we only really need to know what we've covered before, however there's more we can do. Firstly, it's clear that dividing by a complex number doesn't really have an intuitive way of being explained (yet). Traditionally, division was a set operation to divide a quantity into certain discrete and equal amounts. However, we can't divide a pizza into $2 + 4i$ pieces, so is there a way to make division a little easier to work with.

Consider the value $\frac{5 + 10i}{1-2i}.$ There's not much information we can get from this yet, however being able to do so involves a similar method to rationalising radicals in the denominator of a fraction. For this, we will need to consider the conjugate of the denominator. We will cover the conjugate momentarily, but it's essentially swapping the sign of the imaginary part. For example, the conjugate of $1-2i$ is going to be $1 + 2i$ as all we do is swap the sign of the imaginary value. Using this, we can take our original expression and multiply it by one. That is,

$ \begin{align} \frac{5 + 10i}{1-2i} &= \frac{5 + 10i}{1-2i} \times 1\\ \,\\ &= \frac{5 + 10i}{1-2i} \times \frac{1 +2i}{1+2i}\\ \,\\ &= \frac{(5+10i)(1+2i)}{(1-2i)(1+2i)}\\ \,\\ &= \frac{-15 + 20i}{1^2 + 2^2}\\ \,\\ &= \frac{-15 + 20i}{5}\\ \,\\ &= -3+4i. \end{align} $

Indeed, this will always work with the complex conjugate for every case of this realisation. The denominator is essentially a factored complex difference of two squares, resulting in a real value. This allows us break up this number into real and complex components in the form $z = x+iy,$ which is much easier to work with. It may be important to consider another example of this, so please see the below example if you are unsure.

To realise a complex fraction, multiply the numerator and denominator by the complex conjugate of the denominator; this will make the denominator real.

Example: Realise $\frac{3 - 11i}{3 + 4i}.$

To do this, first consider the complex conjugate of the denominator, that is, $3 - 4i$ (changing the sign of the imaginary value). Then, we multiply the number and denominator by the conjugate. Remember, this is equivalent to multiplying by one.

$ \begin{align} \frac{3 - 11i}{3 + 4i} &= \frac{3 - 11i}{3 + 4i} \times \frac{3 - 4i}{3-4i}\\ \,\\ &= \frac{(3-11i)(3-4i)}{(3+4i)(3-4i)}\\ \,\\ &= \frac{-35-45i}{25}\\ \,\\ &= -\frac{7}{5} - \frac{9}{5}i. \end{align} $

We've observed one use for the complex conjugate, so now let's consider some more properties of them. These properties will be purely algebraic for the time being, so before we look into alternative implications of these, we will simply consider them in algebraic contexts. So, suppose we have a complex number $z = x + iy.$ We denote the complex conjugate of $z$ as $\bar{z} = x - iy.$

Let $z, w \in \mathbb{C}$ (this means that we're letting $z$ and $w$ be complex numbers). Then the following properties hold:

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

$\displaystyle \overline{\left( \frac{z}{w}\right)} = \frac{\bar{z}}{\bar{w}};$

$\overline{z \pm w} = \bar{z} \pm \bar{w}$

$z + \bar{z} = 2 \text{Re}(z);$

$z - \bar{z} = 2 \text{Im}(z);$

$z \times \bar{z} = |z|^2;$

$\text{Re}(z) = \text{Re}(\bar{z})$ and $\text{Im}(z) = - \text{Im}(\bar{z}).$

Notably, if we have $z=x+iy,$ then we denote $x = \text{Re}(z)$ (the real part) and $y = \text{Im}(z)$ (the imaginary part).

We will cover the proofs in order as they appear in the list above. Most of these proofs will algebraically boil down to the same method.

Let $z = x_z + iy_z$ and $w = x_w + iy_w.$ For the time being, we'll denote the conjugate of $z$ as $z^*$ (just because it'll be more convenient to work with).

$ \begin{align} \overline{zw} &= \left[(x_z + iy_z)(x_w + iy_w)\right]^*\\ \,\\ &= \left[(x_zx_w - y_zy_w) + i(x_zy_w + x_wy_z) \right]^*\\ \,\\ &= (x_zx_w - y_zy_w) - i(x_zy_w + x_wy_z)\\ \,\\ &=\left[x_zx_w - (-y_z)(-y_w)\right] + i\left[x_z(-y_w) +x_w(-y_z)\right]\\ \,\\ &= (x_z - iy_z)(x_w - iy_w)\\ \,\\ &= \bar{z}\bar{w}. \end{align} $

We will approach this proof the same way, however it's important to note that this isn't necessarily the best method to prove this. Considering the extent of what we've covered so far, the algebraic proof will be the best one for now. In this proof, we will use our properties that $z\bar{z} = |z|^2.$ While we may not currently understand what exactly $|z|$ means, all you need to know from now is that this number is real. As such, we can also observe that $\overline{nz} = n\bar{z}$ for any real number $n.$ Finally, it is obvious that the conjugate of the conjugate is just the original number as swapping the sign twice restores the original one.

$ \begin{align} \overline{\left(\frac{z}{w} \right)} &= \overline{\left(\frac{z\bar{w}}{w\bar{w}}\right)}\\ \,\\ &= \overline{\left(\frac{z\bar{w}}{|w|^2}\right)}\\ \,\\ &= \frac{1}{|w|^2} \overline{z\bar{w}}\\ \,\\ &= \frac{1}{|w|^2}\bar{z}w\\ \,\\ &= \frac{\bar{z}w}{\bar{w}w}\\ \,\\ &= \frac{\bar{z}}{\bar{w}}. \end{align} $

Let $z = x_z + iy_z$ and $w = x_w + iy_w.$

$ \begin{align} \overline{z \pm w} &= \left[(x_z + iy_z) \pm (x_w + iy_w) \right]^*\\ \,\\ &= \left[(x_z \pm x_w) + i(y_z \pm y_w) \right]^*\\ \,\\ &= (x_z \pm x_w) - i(y_z \pm y_w)\\ \,\\ &= (x_z \pm x_w) + i\left[(-y_z) \pm (-y_w) \right]\\ \,\\ &= \left[x_z + i(-y_z)\right] \pm \left[x_w + i(-y_w) \right]\\ \,\\ &= \bar{z} \pm \bar{w}. \end{align} $

Let $z = x+iy.$

$ \begin{align} z + \bar{z} &= (x + iy) + (x - iy)\\ \,\\ &= 2x\\ \,\\ &= 2 \text{Re}(z). \end{align} $

Let $z = x+iy.$

$ \begin{align} z - \bar{z} &= (x + iy) - (x - iy)\\ \,\\ &= 2y\\ \,\\ &= 2 \text{Im}(z). \end{align} $

Again, we have yet to formally define $|z|,$ so if it helps proceed with this section until you understand what this is and then return to this proof once you understand, however for the time being, we define if $z = x+iy,$ $|z| = \sqrt{x^2 + y^2}.$ So, let $z = x+iy.$

$ \begin{align} z \times \bar{z} &= (x + iy)(x-iy)\\ \,\\ &= x^2 + y^2\\ \,\\ &= |z|^2. \end{align} $

These two are by definition true, I won't write out a full proof for these.

Moving on from this, we will cover one final important technique. At some point, we will end up solving quadratic equations with complex coefficients, like $x^2 + 2x + i = 0.$ Note that our discriminant will be $4 - 4i,$ but we're only aware of the square root of real numbers. So, in order to find the square roots of a complex number, you need to notice that if $z$ represents a square root of $w,$ then $z^2 = w.$ This makes it easier to solve. Have a look at the following example to see this, the example is important as it covers the concept of comparing coefficients.

To find the square roots, let some unknown number represent the square roots and square the expression. Compare coefficients and find values that satisfy our conditions.

Example: Find the square roots of $-3 + 4i.$

Let $z = x+iy$ represent the square roots. So, we have $z^2 = x^2 - y^2 + 2ixy = -3+4i$ as squaring will result in what we're trying to find the square root of. Comparing real and imaginary parts of both sides gives us the system of equations $x^2 - y^2 = -3$ (real part of both sides are equal) and $2xy = 4$ (imaginary parts of both sides).

Now, we have simultaneous equations which we can solve. This can be done by inspection (and will often be worth it to do it this way), however, we will work it out manually to cover more extreme cases. From our imaginary components, we can rearrange to get $x = \frac{2}{y}.$ By the substitution method, we arrive at the expression $x^2 - \frac{4}{x^2} = -3$ and so we need to solve $x^4 + 3x^2 - 4 = 0$ which, even though is a quartic, can still be evaluated with the quadratic formula. We will get $x^2 = -4, 1$ but since $x$ is strictly real, we get $x = \pm 1$ as our only solution. Substituting this value back in gets us the solutions $x = 1, y = 2$ and $x = -1, y=-2.$

Therefore, we conclude that the square roots of $-3 + 4i$ are $1 + 2i$ and $-1-2i.$

Since this is essentially raising a complex number to a power, it may seem like there's not clear correlation between a complex number and raising it to any power, however, this will involve a result that we will cover later on. For now, understand the method of this case and we will revisit this in the future.

Geometric Representations of Complex Numbers

From our previous representation of complex numbers, it may have been obvious that we were linking our complex numbers to a set of coordinates. When we think of representing numbers on a graph, we've previously focused on a number line. Since we've only worked with real numbers, we can only really work in this case, however we need to work with complex numbers on a graph as well. So, what we can do is we can define a new dimension. Following from our Cartesian coordinates, we can let the number line remain the same as the $x$-axis. This will be the path of real numbers. Furthermore, we will define a new dimension for the imaginary part. The $y$-axis will be the path of imaginary numbers. Since this occurs when $x = 0,$ it's clear that the number will only be imaginary with no real component. We can let our complex number $z = x+iy$ have respective coordinates on this new graph as $(x,y).$ Consider the examples of complex numbers below.

While this definition allows us to extend our number line to complex numbers, we will come to find that there is a lot more to it than it first may seem. With what we have so far, we can already observe some important features, like for example, complex conjugate will flip across the $x$-axis. However, before we move on to viewing some geometric properties, we should first look at an alternative way of representing complex numbers. We can also represent a complex number in terms of polar coordinates. So, we define the distance from the origin to the complex number as being the modulus and the angle it makes with the real axis as the argument. The argument starts at zero along the positive real axis. Anticlockwise rotations increase the angle and clockwise decrease the angle. We define the principle argument as the argument of the complex number between $-\pi$ and $\pi.$

In terms of the notations, we generally write the modulus of $z$ as $|z|$ and the argument as $\text{arg}(z)$ (for the principle argument, we use $\text{Arg}(z),$). In the diagram below, we have represented the modulus as $|z|$ and $\text{Arg}(z) = \theta.$

Now that we've established these forms of representing a complex number visually, it's important to be able to convert between the two forms. Before we consider an example, we'll briefly unpack exactly how we can do this generally. Representing a complex number in the form $z = x+iy$ is considered the Cartesian form of a complex number. When it comes to the polar representation, we denote it as $z = r \angle \theta,$ where $r = |z|$ and $\theta = \text{Arg}(z).$ We call this form Mod-Arg Form (or polar form). Please note that this notation is not commonly used anymore as you'll come to find there's a better way of representing them, however for the sake of this section, it'll be easier to represent it as such.

We can observe that our coordinates can be represented as a right triangle similar to that of a parametric form of a circle with radius $|z|.$ Understanding this gives us a clear way of identifying a relation between the two forms. One of the most important is that $z = x + iy = r \angle \theta$ where $r = |z| = \sqrt{x^2 + y^2}$ (by Pythagoras' theorem) and $\theta = \text{Arg}(z) = \arctan\left(\frac{y}{x}\right)$ (remember that $\text{Arg}$ is strictly talking about the principle argument). This correlation is represented below, so consider an example to help tie this all together.

Example: Convert $-1 + \sqrt{3}i$ to Mod-Arg form and $\displaystyle 2 \angle \frac{\pi}{3}$ to Cartesian form.

Firstly, we notice from our diagram above that we can consider our first conversion.

By the basic properties of right triangles, we can evaluate $\displaystyle \tan \theta = -\sqrt{3} \Rightarrow \theta = \frac{2\pi}{3}.$ Similarly, with Pythagoras' theorem, we observe that $|z| = \sqrt{1^2 + \sqrt{3}^2} = 2.$ Therefore, we have $\displaystyle -1 + 3i = 2 \angle \frac{2\pi}{3}.$

Moving onto the next case, we will now consider going the other way. If it helps, we will draw up another triangle with the information given.

Using the basic understanding of right angled trigonometry, we can deduce that the $x$ coordinate of the point along the triangle will be $1$ and the $y$ coordinate will be $\sqrt{3}.$ When doing this, make sure to pay attention to the quadrant the sides fall in to determine the signs of the coordinates. So, we finally have that $\displaystyle 2 \angle \frac{\pi}{3} = 1 + \sqrt{3}i.$

Generally, it should be clear that if we're given the argument and the modulus of a complex number, we can find the Cartesian equation as $z = r(\cos \theta + i \sin \theta)$ where $r = |z|$ and $\theta = \text{Arg}(z).$ This notation will prove to be useful in the future, however there's an interesting relationship involving three of mathematics' biggest identities which will make it even easier for us.

Euler's Form of a Complex Number

You may be familiar with the famous identity $e^{i \pi} = -1.$ In this part, we'll be exploring exactly why this is. This form is essentially a different (and arguably better) way of representing complex numbers in polar form.

Define a complex number with $|z| = r$ and $\text{Arg}(z) = \theta,$ then Euler's identity states that $r(\cos \theta + i \sin \theta) = re^{i\theta}.$

Woah! This identity is insanely clean. Working with exponentials are much easier than working with trigonometric functions. This form will indeed prove to be quite useful moving forward. Below are a set of several different proofs for this. These proofs will involve only the case of $\cos \theta + i \sin \theta,$ which is a complex number with a unit modulus. This is the key part of the identity and the link between the two. Now, before we move onto the proofs, it's important to note that the first is the only proof I'd expect someone at this level to understand. The next one are harder, so if you are a more advanced reader then it may be worth having a look.

We aim to prove that $\cos \theta + i \sin \theta = e^{i\theta}.$ So, consider the function $f(\theta) = e^{-i \theta}(\cos \theta + i \sin \theta).$ Differentiating this gives us $$f'(\theta) = -ie^{-i \theta}\cos \theta - e^{-i \theta}\sin \theta + ie^{-i \theta}\cos \theta + e^{-i \theta}\sin \theta = 0.$$ This implies that our original function is a constant, so that means we can take $f(\theta) = f(0) = 1.$ So, we have $$e^{-i\theta}(\cos \theta + i\sin \theta) = 1$$ and thus we arrive at $e^{i \theta} = \cos \theta + i\sin \theta.$

To reiterate, the following proofs are for more advanced readers. Don't stress if you are not sure about these.

Proof One: Series Expansion

For this proof, we will assume you are aware of the Taylor series expansions of $e^x,$ $\sin x$ and $\cos x.$ Specifically, $$\sin x = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n+1},$$ $$\cos x = \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}x^{2n},$$ $$e^x = \sum_{n=0}^\infty \frac{x^n}{n!}.$$ So, with these, we can work from one side to the other.

\begin{align} \cos x + i \sin x &= \left[\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}x^{2n}\right] + i\left[\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n+1}\right]\\ \,\\ &= \sum_{n=0}^\infty \left[\frac{i^{2n}}{(2n)!}x^{2n} + \frac{i^{2n+1}}{(2n+1)!}x^{2n+1}\right]\\ \,\\ &= \sum_{n=0}^\infty \frac{(ix)^n}{n!}\\ \,\\ &= e^{ix}. \end{align}

Proof Two: Particle Argument

Firstly, let's define a particle's displacement on an Argand diagram as $x(t) = e^{it}.$ The velocity of this is hence $\dot{x}(t) = ie^{it}.$ Geometrically, this means that our position vector is going to have the same magnitude as the displacement, just rotated anticlockwise by an angle of $\displaystyle \frac{\pi}{2}.$ Interestingly, this implies that the motion is circular around the unit circle. Coincidentally, our expression $\cos \theta + i\sin \theta$ is the parametric form of this equation of motion. This proof is only outlined here, however it's an interesting proof for this theorem that may make it easier for Physics students to understand.