## UCT MAM1000 lecture notes part 31 – complex numbers part ix

When we were playing around with partial fractions we appeared to make a bit of an assumption which was that the only forms that we had to deal with in the denominator of a fraction could always be written as a factor of either linear parts ( $a+b x$) or quadratic parts which we could not factor into linear parts ( $ax^2+bx+x$) where $b^2-4ac<0$, and of course multiple powers of these, for instance we could have terms like $(a+bx)^3$ in the denominator. How do we know that we can always split a polynomial up into these factors where the coefficients are real? Couldn’t it be for instance that if I gave you a cubic polynomial that all the roots were complex and so I couldn’t factor it in a way that every factor came out with real coefficients? It turns out that the answer is no, but we need a couple more ingredients to prove this.…

## UCT MAM1000 lectures notes part 30 – complex numbers part viii

So, we can do basic algebra with complex numbers, take powers of them, and apply trig and exponential functions to them. There isn’t much left that we might want to do, but taking powers of them is very important and also pretty easy. Let’s say we wanted to calculate the solution to the equation: $z^2=-1$

Well, we know what $z$ is for this, because that’s what got us into this mess in the first place! We know that there are two solutions and they are $\pm i$. That is to say that if you multiply either of these numbers by themselves, you get -1, by definition. How about $z^2=i$. The first thing to do whenever you have to take the root of a complex number, or a real number (here the second root of $i$) is to convert that number into modulus argument form – it will be infinitely easier like that.…

## UCT MAM1000 lecture notes part 29 – complex numbers part vii

So, we know how to take the exponential of any complex number now. We do it by converting the exponential into the exponential of the real and imaginary parts separately, and then use the relationship between $e^{ia}$ and the $\cos$ and $\sin$ functions to write everything in terms of functions of real numbers, which we know how to deal with. How about the trigonometric functions applied to complex numbers? Well, we have a pretty good hint already from how we got from the exponential of complex numbers to trigonometric functions of real numbers. In fact we’re just going to give the answer, but you can work it out using Taylor series as well. For a complex number $z$: $\cos z=\frac{e^{iz}+e^{-iz}}{2}$ $\sin z=\frac{e^{iz}-e^{-iz}}{2i}$

The first thing to check is that this is true when $z$ is a real number. It looks pretty strange at first site, especially the definition of $\sin$ because there’s an $i$ sticking out in the denominator like a sore thumb!…

## UCT MAM1000 lecture notes part 28 – complex numbers part vi

So last time we discovered that there was this amazing link between exponentials and trigonometric functions where the bridge between them was precisely complex numbers.

We now know how to take the exponential of any complex number and it is given by the exponential of a real number and a sum of terms which contain trig functions of real numbers: $e^{a+ib}=e^a(\cos b+i\sin b)$

We can see also that now we have a third way to write a complex number. If you have a number in modulus argument form like: $r(\cos\theta+i\sin\theta)$

Then we can also write this as $re^{i\theta}$. This is an alternative way of writing the modulus argument form.

In this form it also becomes more obvious that moduli multiply and arguments add. If we have two complex numbers: $z=|z|e^{i\theta}$ and $w=|w|e^{i\phi}$ Then: $zw=|z||w|e^{i(\theta+\phi)}$

The fact that we can now take the exponential of any complex number is very powerful. The point is that in order to calculate this function, all we need to be able to do is to take exponentials and trig functions of real numbers, and that we can do.…

## Plotting functions of complex numbers: Not examinable

Just to get a bit of a picture of what taking a function of a complex number means, we can play a bit of a game (I use this term in the loosest sense). Normally we think of functions as going from a real number to another real number. $\sin(x)$ takes a real number $x$ and gives you another real number. We can plot this on a graph by plotting a two dimensional set of data which tells you about the value that $\sin(x)$ takes for every $x$ along the real line. We are very used to this idea of a function. However, a function of a complex number is more difficult to visualise.

Complex numbers themselves live in 2 dimensions (they have a real part and an imaginary part) and when you apply a function to them, very often the result is another complex number which also lives in a 2 dimensional space.…

## UCT MAM1000 lecture notes part 27 – complex numbers part v

We’re about to make one of the most profound links that we will obtain through complex numbers. This is going to show how complex numbers are a bridge between different areas that you already know about, but never knew had anything to do with one another.

We know about exponential functions and how they have very special properties related to their derivatives. $e^x$ is a function which is practically defined as the function which is equal to its derivative. We also know that exponential functions tell us about growth, and we will see this in more detail when we come on to differential equations.

We know that trigonometric functions are to do with triangles, and circles, and angles and they tend to be periodic. They tell us how things vary in a way where they come back to where they started after some time.

Exponential functions and trigonometric functions couldn’t really look much more different if they tried.…

## UCT MAM1000 lecture notes part 26 – complex numbers part iv

OK, so we saw something pretty interesting last time when we multiplied together complex numbers using the modulus argument form.

Remember that for two complex numbers which we will write as $z_1=r_1(\cos\theta_1+\sin\theta_1 i)$ and $z_2=r_2(\cos\theta_2+\sin\theta_2 i)$, where $r_i$ are the moduli, and $\theta_i$ are the arguments of $z_i$. If we multiply them together then we get: $z_1 z_2=r_1 r_2 (\cos(\theta_1+\theta_2)+i\sin(\theta_1+\theta_2))$

Well, what would happen if the two complex numbers were the same? ie. if we have $z=r(\cos\theta+i\sin\theta)$ and we want $z^2$?

Well, then clearly: $z^2=r^2(\cos 2\theta+i\sin 2\theta)$.

What if we then multiplied this by $z$ one more time: $z^3=z^2 z=r^3(\cos (2\theta+\theta)+i\sin(2\theta+\theta))=r^3(\cos 3\theta+i\sin 3\theta)$

hmm, do we already see a pattern emerging? Let’s say that we have a complex number with modulus 1. Complex numbers of the form: $z=\cos\theta+i\sin\theta$

Are clearly modulus 1. We know that the modulus is the square root of the sum of the squares of the real and imaginary parts of a complex numbers so $|z|=\sqrt{\cos^2\theta+\sin^2\theta}=1$.

ok, so how about if we have $z^n$ where $n$ is an integer?…

## UCT MAM1000 lecture notes part 25 – complex numbers part iii

So we saw last time that we can take a complex number and put it in a 2 dimensional plane called the complex plane, where its horizontal distance from the origin is given by its real part, and the vertical distance from the origin is given by its imaginary part. We can thus think of the real and imaginary parts as the Cartesian coordinates of that point.

It turns out that there is another way to represent a complex number, but rather than using the real and imaginary parts to specify it, we will use two other pieces of information.

If I tell you that a complex number is a distance $|z|$ away from the origin in the complex plane, then this leaves you with a whole circle of possibilities. All the points on the circle of radius $|z|$ about the origin are the same distance from the origin. But if I also give you an angle subtended between the x-axis and the line joining the complex number and the origin, read anti-clockwise from the x-axis, this will completely pin down the point in the complex plane.…

## UCT MAM1000 lecture notes part 23 – Thursday August 20th

Complex numbers

These will be an addition to the notes already on Vula on complex numbers. Please refer to that document as well as I will be taking a slightly alternative approach on occasion.

A philosophical detour

Before we get on to talking about imaginary numbers and complex numbers, let’s try and break down our preconceptions about numbers in general. We look at the world around us and see many things which we categorise. We see a computer, a piece of paper, we see other people, we see our hands. These are labels that we use to categorise the world around us, but these objects seem very physical and very real. We rarely question their existence, though if one wants to take the Cartesian view, we should also question the reality we are in. We are not going to go that far, but let’s try and ask about the existence of numbers.…

## UCT MAM1000 lecture notes part 22 – Wednesday 19th of August

So, last time we looked at the Maclaurin expression for $e^x$. The exponential function was particular easy because its derivative is equal to the function itself every time. Let’s look at a slightly more involved example where this is not true: $f(x)=\sin x$ about $x=0$. Again, we start with the table of derivatives: $\left( \begin{array}{ccc} \text{ i} & f^{(i)}\text{(x)} & f^{(i)}\text{(0)} \\ 0 & \sin (x) & 0 \\ 1 & \cos (x) & 1 \\ 2 & -\sin (x) & 0 \\ 3 & -\cos (x) & -1 \\ 4 & \sin (x) & 0 \\ 5 & \cos (x) & 1 \\ \end{array} \right)$

Now the values of the derivatives are not always the same. They are zero every other term, and they change in sign when they are not zero. This leads to a very elegant expression for the $\sin$ function expanded around $x=0$: $\sin x\approx\sum_{i=0}^n \frac{(-1)^ix^{2i+1}}{(2i+1)!}$

An important point is that here the terms get smaller and smaller as you take more and more of them, so if, for instance, you want to know the value of $\sin 2.4$ you can plug it into the right hand side, take a finite number of terms and you will get an approximation: $\sum_{i=0}^n \frac{(-1)^i2.4^{2i+1}}{(2i+1)!}$

The higher value you choose for $n$ the more accurate will be your answer, but we can see that we can now, in theory calculate the $\sin$ of any number with pen and paper, so long as you have enough patience and will-power.…