4
$\begingroup$

Students often make the mistake of writing the following:

$$\frac{1}{a+b} = \frac{1}{a}+\frac{1}{b}$$

However, after doing a bit of algebra, it turns out that the above has solutions defined by:

$$a=be^{i\frac{2\pi}{3}+2n\pi i},\ n\in\mathbb{Z}$$ and: $$a=be^{i\frac{4\pi}{3}+2n\pi i},\ n\in\mathbb{Z}.$$

Using the original equation, it can be shown that:

$$(e^{i\frac{2\pi}{3}}+1)^{-1} = e^{-i\frac{2\pi}{3}}+1$$ $$(e^{i\frac{4\pi}{3}}+1)^{-1} = e^{-i\frac{4\pi}{3}}+1$$

Are there any interesting applications of these identities?

$\endgroup$
  • $\begingroup$ .. if a student knows what is the exponential form of complex numbers that he never made such mistake with those fractions. $\endgroup$ – Leox Jan 9 '14 at 20:10
  • 1
    $\begingroup$ I was explaining the motivation behind the identities/equalities. $\endgroup$ – chs21259 Jan 9 '14 at 20:12
  • 1
    $\begingroup$ There is no need to the $+2n\pi \operatorname{i}$ in your expressions. You're just adding a full turn, which doesn't change the number. $\endgroup$ – Fly by Night Jan 9 '14 at 20:35
  • 1
    $\begingroup$ I don't know... Would you consider this to be interesting ? :-) $\endgroup$ – Lucian Jan 10 '14 at 5:32
  • $\begingroup$ @Lucian That is just the kind of problem I was looking for. How would I go about applying the above to a problem like that? $\endgroup$ – chs21259 Jan 10 '14 at 20:45
3
$\begingroup$

Let $\omega = e^{i\frac{2\pi}{3}}$, a cube root of unity. Since $(\omega - 1)(\omega^2 + \omega + 1) = \omega^3 - 1 = 0$, and $\omega \ne 1$, $$ \omega^2 + \omega + 1 = 0, $$ and so we obtain the identites: $$ \omega + 1 = - \omega^2 = - \omega^{-1} $$ and $$ -\omega = \omega^2 + 1 = \omega^{-1} + 1 $$ Therefore, $$ \left( \omega + 1 \right)^{-1} = \omega^{-1} + 1. $$

$\endgroup$
  • $\begingroup$ Very cool. Thanks! $\endgroup$ – chs21259 Jan 9 '14 at 20:47
2
$\begingroup$

The cube roots of unity are special in various ways - because they are quadratic they are constructible in the Euclidean sense (just as an equilateral triangle is constructible). They also have special properties because of their quadratic character (Hardy & Wright "An Introduction to the Theory of Numbers" gives the old-fashioned version of this - and surprising depth for an old-fashioned treatment). The cube roots of unity emerge from the equation you give as perhaps initially unexpected. But once identified, the quadratic field they define is well known.

$\endgroup$
  • $\begingroup$ Thank you for posting a constructive answer. I'll look into that $\endgroup$ – chs21259 Jan 9 '14 at 20:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.