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}. There was a whole roadblock in terms of the operations which you could apply to numbers and there were frequently equations which simply didn’t have solutions. Had you been asked to find any of:

\sin x=5



you simply wouldn’t have been able to continue.

What we have done in the last section is to open up the possibilities of all of these equations which were previously met with confusion an frustration. We have done this by introducing a simple new ingredient – a number whose square is negative and suddenly all of these problems which were previously intractable are now…well…tractable.

This might seem like just fun and games, but complex numbers play an enormous role in helping us in large areas of mathematics. They lead us to solutions which otherwise remain hidden and frequently give us enormous shortcuts in our calculational landscape.

Over the last few sections we have seen lots of tools and techniques which allow us to solve interesting problems. For now, you should familiarise yourself with these tools and manipulations, both through the notes which I have put up, and also through the extra complex numbers notes in the resource book. Please, please read these – this is not a nice addition, but there are a few ways of doing things which I have not explicitly discussed but you should know. I have explained everything which I believe may not be easy to understand if you are reading it on your own, but there are a few important rules which you should go through yourself. If you have any questions, be sure to email me and I will be happy to explain.

Here I will just put in a few more examples which might help to add understanding to your complex numbers comprehension.


Solve e^z=1+i for z.



We start by writing z=a+bi and 1+i in the exponential modulus argument form, also known as Euler form. This renders the equation into:

e^{a+bi}=\sqrt{2}e^{i\frac{\pi}{4}+i2n\pi}\,\, \, n\in{\mathbb Z}

The left hand side can be split into e^{a+bi}=e^ae^{bi} and now we simply equate the modulus part with the argument part:


e^{ib}=e^{i\frac{\pi}{4}+i2n\pi}\,\, \, n\in{\mathbb Z}

such that a=\ln\sqrt{2} and b=\frac{\pi}{4}+2n\pi with n\in{\mathbb Z}, so the solution is z=\ln\sqrt{2}+i\left(\frac{\pi}{4}+2n\pi\right) with n\in{\mathbb Z}


Find the real part of (1-\sqrt{3}i)^{31}.



We will use de Moivre for this. First we write this in the form r\cos\theta+ir\sin\theta. In this case \theta=\arctan\left(-\sqrt{3}\right)=-\arctan\sqrt{3}=-\frac{\pi}{3}+2n\pi and r=\sqrt{1+\sqrt{3}^2}=2. So


So using de Moivre:


But we know that \cos and \sin are periodic with period 2\pi and so we can alter the argument of these functions by factors of 2n\pi and the answer will remain the same. If we subtract 10\pi from \frac{31\pi}{3} we get \frac{pi}{3} and thus \cos(31\frac{\pi}{3})=\cos(\frac{\pi}{3}). The real part of (1-\sqrt{3}i)^{31} is thus:


Solve \cosh z=-3.


We know that we can write \cosh in terms of exponents and so we have:


We proceed by multiplying through by 2 and by e^z and writing the whole expression as a quadratic equation:

e^{2z}+6 e^z+1=0

We can now use the quadratic formula to give:


Writing z=a+bi and writing the right hand side in modulus argument form we get:

e^ae^{bi}=(3\pm\sqrt{8})e^{i\pi+2 n\pi i}\,\,\,\, n\in{\mathbb Z}

Note that we have converted -1 into a phase, because the magnitude is always a real number. Check for yourself that the above is true. Now we simply solve for a and b as before and get:

z=\ln\left(3\pm\sqrt{8}\right)+i (\pi+2n\pi)\,\,\, n\in{\mathbb Z}

Finally, let’s find the 6 roots of z^6+\sqrt{2}z^3+1=0.

We see that this is a quadratic in z^3 and so we can write:

z^3=\frac{-\sqrt{2}\pm \sqrt{2-4}}{2}=\frac{-\sqrt{2}\pm i\sqrt{2}}{2}=\frac{1}{\sqrt{2}}\left(-1\pm i \right)

We now write this in Euler form to get:

z^3=e^{\pm\frac{3\pi i}{4}+2n\pi i},\,\, n\in{\mathbb Z}

and so:

z=e^{\pm \frac{i\pi}{4}+\frac{2n\pi i}{3}}\,\,\, n\in{\mathbb Z}

These six solutions are given by the 3 conjugate pairs:

\begin{array}{cc}  z=\frac{1\pm i}{\sqrt{2}} \\  z=-\frac{1}{2} \sqrt{2+\sqrt{3}}\pm\frac{1}{4} i \left(\sqrt{2}-\sqrt{6}\right) \\  z=\frac{\sqrt{2-\sqrt{3}}}{2}\pm\frac{1}{2} i \sqrt{2+\sqrt{3}}  \end{array}

ok, so these are just a very few examples, but the techniques here cover a good deal of what we’ve done. You should become as familiar with complex numbers as you are with rational numbers, and negative numbers and indeed the integers which was were the story all started…

How clear is this post?