Complex numbers this week bring us to the end of our journey through the number systems. I found this much easier going, and I hope the students did too.
Complex numbers provided conceptual difficulties for mathematicians, who were reluctant to admit them as genuine numbers, hence the pejorative term “imaginary”. This is not dissimilar to other terms applied to new numbers as they were introduced: “negative”, “irrational”, but also to terms like “ideal” (and even “infinite”). Complex numbers were admitted as a tool for calculation – indeed, they are needed for solving cubic equations – but were for some time rejected as actual solutions to the equations. One of the students, Emma Kelly, sent me this cartoon which illustrates the point.
Despite this, the complex numbers are not hard to do calculations with; everything is very hands-on, just do what you are used to, including expanding brackets – the distributive law – and remember that i2 = −1.
- How do you show that every positive real number has an nth root? You have to observe that there is an algorithm for finding the decimal expansion of the nth root of a given number, and use a completeness property of the real numbers (such as the Principle of the Supremum) to guarantee that you have actually defined a number.
- What about the nth root of any complex number? By De Moivre’s theorem, we simply take the nth root of the modulus, and divide the argument by n, to find the modulus and argument of the root. With little extra effort we can see that there are n distinct nth roots (if the original number isn’t zero), which in the Argand diagram form the vertices of a regular n-gon with centre at the origin.
As an added bonus, the proof of De Moivre’s Law is a nice easy example of a proof by induction.
eiπ = −1
The complex numbers bring us to Euler’s formula, one of the most famous in mathematics, combining the negative and imaginary units and the two most famous mathematical constants. So of course you would like to see a proof!
The reason I can’t show you one is the same as the reason I couldn’t prove that 0.999… = 1 last week. The problem then was “What does an infinite decimal mean?”, and the answer was “It is a limit, and we have to define the limit of an infinite sum, and prove that in this case the limit exists and is equal to 1″. This time, the problem is that we haven’t defined what it means to raise a number to an imaginary power.
I think the best answer to this question is one that can’t really be given to a student who hasn’t yet come across complex analysis. This is a very powerful subject, which is the basis for quantum theory (among many other things), and the modern world couldn’t exist without it. (That is what I call impact, though I don’t expect funding committees to be impressed.) In order for complex analysis to make sense, it is necessary that a power series expansion of a function which is valid for real values holds also for non-real complex values. Now if we apply this principle to the power series for the exponential function, substitute an imaginary number ix into the series, and separate into real and imaginary parts, we obtain
eix = cos x + i sin x.
Now putting x = π into this gives Euler’s formula.
Is this satisfactory? The alternative approach, it seems to me, is to take the displayed formula as a definition; while this gives all the consistency we want, it seems a bit unmotivated to me.
The real miracle is that, while the rule for adding complex numbers (“add the real parts and add the imaginary parts”) is straightforward, the rule for multiplying is less so (though easily derived from the requirement that we want the usual laws to hold and also we want i2 = −1). However, when we use polar coordinates, that is, put our complex numbers into modulus-argument form, we see that the rule for multiplication is simply “multiply the moduli and add the arguments”. This is what makes the magic work, and it follows from the addition laws for sine and cosine. So it is the properties of these functions that give us what we want, just as surely as it is in my explanation in terms of power series.
Solving the cubic
Solving equations provides some of the most romantic narratives in the history of mathematics. I didn’t include this in the lectures, but in the supplementary material I showed how to solve a cubic using arithmetic operations and the extraction of square roots and cube roots. It is not too hard, and you can probably fill in the details, even if you haven’t seen it before. So take a general cubic equation ax3+bx2+cx+d = 0.
- Dividing through by the leading coefficient, we may assume that a = 1.
- “Completing the cube”, we may assume that b = 0.
- Now we use the remarkable identity
(x+y+z)(x+ωy+ω2z)(x+ω2y+ωz) = x3+y3+z3−3xyz,
where ω is a cube root of unity. If we could choose y and z such that y3+z3 = d and −3yz = c, then we would have the solution. Now this can easily be done by observing that y3 and z3 are the roots of a quadratic whose coefficients we know. So solve the quadratic, extract a cube root (one is all we need, since we can find z if we know y), and we are done!