Ask Uncle Colin: an irrational power

Dear Uncle Colin,

How would you work out something like $(-1)^\pi$?

Doesn’t Everyone Miss Out Irrational Values Reckoning Exponents?

Hi, DEMOIVRE, and thanks for your message! In the words of the politician: difficult, difficult, lemon difficult. One the one hand there is one answer that seems sensible - but on the other, one can quite reasonably argue that there are infinitely many complex numbers on the unit circle that are $\pi$th roots of -1.

Principal value

The standard way of finding the principal value of $z^n$ - for integer $n$ - is to use de Moivre’s theorem1. If you express $z$ as $\cos(\theta) + \i \sin(\theta)$, then $z^n$ works out to be $\cos(n\theta) + \i \sin(n \theta)$ - which is a really neat result that has all sorts of lovely consequences for trigonometric identities and suchlike. But we’ll gloss over those.

For rational values of $q$, things get more complicated for $z^q$. The problem is that angles don’t have a unique representation – for example, you can write -1 as $\cos(\pi) + \i \sin(\pi)$ or $\cos(-\pi) + \i\sin(-\pi)$ - among infinitely many others. If you multiplied any of those by a given integer, you get the same result; when you multiply by a rational number, that’s not the case. For example, multiplying the angle by $\frac{1}{2}$ in some cases gives something coterminal2 to $\piby 2$, and in others to $-\piby 2$. Square rooting in the complex plane gives two answers - and we can pick one of them to be the principal value. By convention, the square root of -1 is $\i$ rather than $-\i$.

So, adopting that convention, the principal value of $(-1)^\pi$ would be $\cos(\pi^2) + \i \sin(\pi^2) \approx -0.902 - 0.430\i$.

But wait!

The thing with rational powers in the form $\frac{p}{q}$ is, they divide the possible arguments into equivalence classes modulo $2\pi$ - and there are at most $q$ of those - as you saw, the square root divided all of the possible angles into two classes of coterminal angles.

The trouble is, irrational numbers don’t do that. Irrational numbers give infinitely many possible values for $\theta$. (To see this, imagine representing $\pi$ as successively more accurate fractions. If you approximated it as $\frac{22}{7}$, you would have seven solutions. If you went for $\frac{355}{113}$, you would have 113. If you picked another more accurate approximation, you would have more - and more - and more. And there’s always a more accurate approximation, so there are always more solutions.


So, as far as I’m concerned, the answer to this is “run away and hide until they ask you something nicer”. There is a principal solution, but infinitely many others3.

I hope that’s some sort of help!

- Uncle Colin

* Edited 2019-05-08 to fix some LaTeX.

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. Hey! That’s your name! []
  2. i.e., the angle ends in the same place. $\piby 3$ is coterminal with $\frac{7}{3}\pi$. []
  3. It’s not the case that every $z$ on the unit circle is a solution - but that’s a head-scratcher for another day. []

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