## A tricky complex numbers problem

The question is as follows:

If $\frac{\pi}{6}\in arg(z+a)$ and $\frac{2\pi}{3}\in arg(z-a)$ and $a\in \mathbb{R}$, find $z$.

So, let’s think about the information given and what we are trying to find. We want to find the complex number $z$ which satisfies this slightly strange set of constraints, and the constraints are given in terms of $z$ and $a$. So, by the looks of things, the answer will depend on $a$ and so the final expression should be a function of $a$.

Now let’s explore the constraints. In fact, let’s simply take $z+a$ and $z-a$ as two complex numbers, but importantly two numbers which differ only by a real number $2a$, so wherever they lie in the complex plane, they have the same imaginary part and differ only by an real part.

Now, the constraints are about the arguments of the two complex numbers. It doesn’t tell us anything about the magnitude of the numbers, so all the information tells us is the direction are in relation to the origin.…

## Pi and the Mandelbrot set

For those studying calculus, can you see what Pi has to do with the Mandelbrot set? I’ll give you a hint, it has to do with certain integrals which you have almost certainly studied in class. We will have a post explaining exactly why this number pops up in this particular place, soon!

If you don’t know about the Mandelbrot set, take a look here, then have a look at the video above.

 How clear is this post?

## UCT MAM1000 lecture notes: More complex numbers practice

I’ve been asked a few times for more practice questions on complex numbers. This is where Wolfram Alpha can be your friend (like it’s not already!).

I’ll just give a few examples of questions from the tut on complex numbers which you could have solved using Wolfram Alpha, and from this you will be able to set up your own questions.

For instance, question 48 c) Find the roots of $z^5=1+\sqrt(3) I$ can be solved in Wolfram Alpha with the command:

Solve[z^5==1+sqrt[3]I,z]

Moreover it will solve this for you, give you the five roots and plot them in the complex plane. So now you can come up with any root question you can possibly think of. There’s an infinite number of questions to start you off. You can thank me later!

If you want to convert between the trigonometric form and the exponential form, you can use the two commands:

TrigToExp[Sin[x]+2 I Cos[x]]

ExpToTrig[Exp[I z+3]]

Though remember the definition of the hypergeometric trig functions from a previous tut.…

## UCT MAM1000 lecture notes part x – summary

When you were a child you first learnt how to count. You learnt the relationship between a number and the quantity of objects around you. You quickly learnt to manipulate these numbers, you could add them, and you could multiply them, and you could subtract them. Subtracting them brought you to negative integers. Then you learnt how to divide them and this took you to fractions and thus the rational numbers. You could then manipulate these new numbers with the same operations.

At school you learnt about powers of numbers, and square roots of numbers, and you learnt about equations and how you could solve them to find the numbers which satisfied the equation. There were a few rules which acted like dead-ends however. You could never take the square root of a negative number. In fact any fractional power of a negative number should have filled you with trepidation (and possibly a frisson of excitement), though sometimes you saw that there were sneaky solutions to things like $(-1)^\frac{1}{3}$.…

## Complexity from complex numbers – The beauty of the Mandelbrot set

We are about to show that you can get incredible structure from the simplest of algorithms when we use complex numbers.

The equation we are going to look at is an iterative equation:

$z_{i+1}=z_i^2+C$

with $z_0=0$. You simply get the next $z_i$ from plugging in the previous one, squaring it and adding a number $C$. I’m going to give you a value for $C$, then you’re going to iterate this equation and see what happens. For instance, if I give you the number $C=3$:

$\left( \begin{array}{ccc} i & z_i^2+C & \left| z_{i+1}\right| \\ 1 & 0^2+3 & 3 \\ 2 & 3^2+3 & 12 \\ 3 & 12^2+3 & 147 \\ 4 & 147^2+3 & 21612 \\ \end{array} \right)$

You can see that this number is just going to keep on increasing without end if we keep applying the algorithm. How about a smaller number, let’s say $C=0.1$:

$\left( \begin{array}{ccc} i & z_i^2+C & \left| z_{i+1}\right| \\ 1 & 0^2+0.1 & 0.1 \\ 2 & 0.1^2+0.1 & 0.1121 \\ 3 & 0.1121^2+0.1 & 0.112566 \\ 4 & 0.112566^2+0.1 & 0.112671 \\ \end{array} \right)$

It looks like this is tending to some value. In fact it has come to a fixed point where $z=z^2+0.1$. There are actually two solutions to this equation but one of them is 0.112702 which is where we are tending towards.…

## 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.…