# $\cos(72º)$, revisited again: De Moivre’s Theorem

In previous articles, I've looked at how to find $\cos(72º)$ using some nasty algebra and some comparatively nice geometry. In this one, inspired by @ImMisterAl, I try some nicer - although quite literally complex - geometry.

### De Moivre's Theorem

I'm going to assume you're ok with complex numbers. If you're not, go and read up on them. They're cool. I'll wait.

One of the many nice properties of complex numbers is that there are two ways to write them: in rectangular form (such as $z=a+bi$) and in polar form, something like $z=R e^{i\theta}$, where $a = R \sin(\theta)$ and $b= R \cos(\theta)$1

This property means we can use exponent properties on complex numbers: quite clearly, if $z=R e^{i\theta}$, then $z^2 = e^{i(2\theta)}$ and more generally $z^n = e^{i(n\theta)}$ (for integer $n$). Incidentally, this is a very nice way of proving the double angle formulas if you have nothing better to do.

This idea, known as De Moivre's theorem, is a useful link between algebra and trigonometry.

### Roots of unity

In particular, if we think about the angle of 72º, or $\frac{2}{5}\pi$ in proper units, we can see that five of them together make up a whole circle. So, if we let $Z = \cos\br{\frac{2}{5}\pi}+i \sin\br{\frac{2}{5}i}$, or $e^{i\br{\frac{2}{5}\pi}}$ it's straightforward to see that $Z^5 = e^{i(2\pi)} = 1$.

So, our complex number $Z$ must satisfy $Z^5 - 1=0$ - which factorises as $(Z-1)(Z^4 + Z^3 + Z^2 + Z + 1)=0$. We know that $Z \ne 1$, so we can focus on $(Z^4 + Z^3 + Z^2 + Z + 1)=0$ instead.

### Conjugates

But wait - there's more! The complex conjugate of a number $z = a+bi$ is $\bar z = a - bi$ - and by looking at the unit circle, it's clear than $Z^3 = \bar Z^2$ and $Z^4 = \bar Z$.

So, our $(Z^4 + Z^3 + Z^2 + Z + 1)=0$ becomes $(\bar Z + \bar Z^2 + Z^2 + Z + 1)=0$.

It's also worth noting that $z + \bar z = 2a$, for any $z$.

For our value of $Z$, we have our '$a$' - the real part - as $\cos\br{\frac{2}{5}\pi}$. For $Z^2$, the real part is the cosine of twice that angle, which is $2a^2 -1$ using the double angle formula.

So, we know that $Z + \bar Z = 2a$ and $Z^2 + \bar Z^2 = 2\br{2a^2 - 1}$. We can substitute these into the bracket to get: $4a^2 - 2 + 2a + 1 = 0$, which simplifies to $4a^2 + 2a -1 =0$.

Now all we need to do is solve it! $a = \frac{-2 \pm \sqrt{4+16}}{8} = \frac{-1 \pm \sqrt{5}}{4}$. We know $a > 0$ because it's in the first quadrant, so the value of $\cos(72º)$ is $\frac{\sqrt{5}-1}{4} \blacksquare$ ## Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

1. That might look familiar if you've been trying to put things in the form $R \sin(x-\alpha)$ at A-level. It's not quite the same thing, but it's the same flavour. []

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