### Related Content Outcomes:

- MEX-N2 Using Complex Numbers
- N2.1: Solving Equations with Complex Numbers

## Terrible Puns, Terrible Maths

Excuse the terrible pun on “dangerous moves” in the title (I thought it was brilliant) but it’s to highlight the all too common mistake made by students (and teachers) of elementary high school mathematics of overlooking necessary assumptions that make a particular theorem or formula work.

In this post, the theorem to explore this particular problem is, as given away by the title, De Moivre’s Theorem.

Many Extension II students know the theorem in the topic of Complex Numbers as this:

\[ (\cos \theta +i \sin \theta)^n = \cos (n\theta) + i \sin (n \theta) \]

So the tempting, and often done, action is to carry out this identity on anything that remotely looks like the left hand side. However, this is where in some cases, that can go wrong!

## Check Your Assumptions

Before we delve more into De Moivre’s Theorem, let’s take a step back first and have a look at this classic party trick to confuse people who don’t check underlying assumptions:

\[\begin{align*}

1 & = \sqrt{1}\\

& = \sqrt{-1 \times -1}\\

& = \sqrt{-1} \times \sqrt{-1}\\

& = \sqrt{-1}^2\\

& = -1

\end{align*}

\]

We’ve just proven \( 1 = -1 \)!

However, this “proof” falls apart when we re-examine the assumptions in using the result we learned in Year 9 Surds:

\[ \sqrt{ab} = \sqrt{a} \times \sqrt{b} \]

This can only be used when \(a > 0\) and \( b > 0 \), which means we are not allowed to use it for negative numbers like in the “proof” that \( 1 = -1\). The reason why this is the case shall be explored later in this post.

So let’s have a look at another “proof” that \( 1 = -1 \), this time using De Moivre’s Theorem incorrectly:

\[\begin{align*}

1 = \cos 0 + i \sin 0 & = \cos 2 \pi + i \sin 2 \pi\\

(\cos 0 + i \sin 0)^\frac{1}{2} & = (\cos 2 \pi + i \sin 2 \pi)^\frac{1}{2}\\

\cos 0 + i \sin 0 & = \cos \pi + i \sin \pi \mbox{ (By De Moivre’s Theorem)}\\

1 & = -1

\end{align*}

\]

What went wrong here?

When we don’t check the cases for which De Moivre’s Theorem is allowed to be applied, contradictions like this arise.

## De Moivre’s Theorem

The Syllabus Glossary states the following for De Moivre’s Theorem:

De Moivre’s theorem states that for all integers \(n\),

Mathematics Extension 2 Stage 6 Syllabus (2017) – Page 44

\[ [r(\cos\theta + i \sin\theta)]^n = r^n(\cos n\theta + i \sin n \theta) \]

This states that the theorem only works for integer values of \(n\) and not fractional powers! As we’ve seen, trying to use De Moivre’s Theorem for fractional powers yields multiple values.

So what is actually happening when we try to use the theorem for rational powers?

## Extending For Rational Powers

Lets have a look at solving multi-angled trigonometric equations for \( x \in [0,2\pi]\) such as:

\[ \sin 3x = \frac{1}{2} \]

We need to add 2 extra revolutions to the 1st revolution solutions, as the division by 3 in the last step brings them back into the domain \( [0,2\pi] \):

\[\begin{align*}

3x & = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{13\pi}{6}, \frac{17\pi}{6}, \frac{25\pi}{6}, \frac{29\pi}{6}\\

x & = \frac{\pi}{18}, \frac{5\pi}{18}, \frac{13\pi}{18}, \frac{17\pi}{18}, \frac{25\pi}{18}, \frac{29\pi}{18}

\end{align*}

\]

In a similar manner, the division process in the application of De Moivre’s Theorem to rational powers would bring arguments that are outside \([0,2\pi]\) back in.

So, the values of \((cos\theta + i \sin \theta )^\frac{1}{n}\) for \( n \in \mathbb{Z} \) can be found by taking \(n\) revolutions of the argument of the complex number, and the division by \(n\) will bring all of them back into \([0,2\pi]\).

Hence:

\[\begin{align*}

(\cos\theta + i \sin\theta)^\frac{1}{n} & = (\cos (2k\pi + \theta) + i \sin(2k\pi + \theta))^\frac{1}{n} \mbox{ (for \(k = 0,1,2,\ldots, n-1\))}\\

& = \cos \frac{2k\pi + \theta}{n} + i \sin \frac{2k\pi + \theta}{n}

\end{align*}

\]

This is how De Moivre’s Theorem is extended for rational indices. We’ve now shown that if the denominator of the index is \(n\), there will be \(n\) values when applying the theorem.

So what happens if the index is irrational (i.e. if it is in \(\mathbb{R}\backslash\mathbb{Q}\))?

As there is no way to divide the interval \([0,2\pi]\) into an irrational number of equally spaced sub-intervals, there will actually be an uncountably infinite number of values!

In fact, when we want to look at complex valued exponents, something as innocent looking as \(1^\pi\) actually has an infinite number of values! One of these values can be found by:

\[\begin{align*}

1^\pi & = (\cos 2\pi + i \sin 2\pi)^\pi\\

& = \cos 2\pi^2 + i \sin 2 \pi^2\\

& \approx 0.6296817253 + 0.7768532196i

\end{align*}

\]

The other values can be found by adjusting the argument to the next multiple of \(2\pi\).

Weird!

There are other complex valued functions that also do not behave the same way their real valued cousins do. For example, complex valued logarithms have multiple values, the index law \((a^m)^n = a^{mn}\) doesn’t actually hold true, solving complex trigonometric equations (such as \(\sin x = 2\)) have solutions, and integrals with complex numbers? Another can of worms altogether!

It’s important to re-examine assumptions when studying any new piece of Mathematics, no matter how basic we may think them to be.

## One More Thing…

Let’s have a look at the complex logarithm. Consider a complex number written in its Exponential form: \(z = re^{i\theta}\).

\[\begin{align*}

e^y & = z\\

e^y & = re^{i\theta}\\

e^y & = re^{i(2k\pi + \theta)} \; k\in\mathbb{Z}\\

y & = \log_e(re^{i(2k\pi + \theta)})\\

y & = \log_e(r) + i(2k\pi + \theta)

\end{align*}

\]

We can see that the complex logarithm is multi-valued! For example:

\[\begin{align*}

\log_e(-1) & = \log_e(1) + i(2k\pi + pi)\; k \in \mathbb{Z}\\

& = (2k+1)\pi i

\end{align*}

\]

Earlier, we asked the question what happens with De Moivre’s Theorem with integer, rational and real exponents. Armed with this knowledge of complex logarithms, one may now answer the question: “What happens with complex exponents?”

Let’s consider finding the value(s) of \(i^i\):

\[\begin{align*}

i^i & = e^{\log_e(i^i)}\\

& = e^{i\log_e(i)}\\

& = e^{i((\log_e(1) + i(2k\pi + \frac{\pi}{2}))}\; k \in \mathbb{Z}\\

& = e^{-(2k\pi + \frac{\pi}{2})}\\

& = 0.2079\ldots, 111.3\ldots, 0.0003882\ldots, 7.249\ldots \times 10^{-7}, \ldots

\end{align*}

\]

First thing to note, is that \(i^i\) is purely Real and the next is that it is multi-valued.

We have seen that the properties of complex numbers and exponents is strange, yet very interesting.

So to conclude: be careful when applying formulae and theorems to situations for which they haven’t been defined for! You’ll find some fascinating relationships when things don’t behave the way you expect.

This is why we should not use i := √(-1), but rather i² = -1.