# Does anything interesting come out of this identity?

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?

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.. if a student knows what is the exponential form of complex numbers that he never made such mistake with those fractions. – Leox Jan 9 '14 at 20:10
I was explaining the motivation behind the identities/equalities. – chs21259 Jan 9 '14 at 20:12
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. – Fly by Night Jan 9 '14 at 20:35
I don't know... Would you consider this to be interesting ? :-) – Lucian Jan 10 '14 at 5:32
@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? – chs21259 Jan 10 '14 at 20:45

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, $$\left( \omega + 1 \right)^{-1} = \omega^{-1} + 1.$$

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Very cool. Thanks! – chs21259 Jan 9 '14 at 20:47

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.

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