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Suppose $a,b,c$ lie in the unit circle of the complex plane and satisfy $a+b+c = 0$. Then, $a,b,c$ form the vertices of an equilateral triangle.

Try

So, I want to show that $|b-a|=|c-a|=|b-c|$. We are given $|a| = |b| = |c| = 1$ and $a+b+c=0$. Since $|z| = z \overline{z} $, we have

$$ (b-a)( \overline{b}-\overline{a} ) = (c-a)( \overline{c} - \overline{a} )$$

which simplifies to

$$ |b| - b \overline{a} - a \overline{b} - |a| = |c| - c \overline{a} - a \overline{c} - |a| $$

Hence,

$$ c \overline{a} + a \overline{c} - b \overline{a} - a \overline{b} = 0$$

$$ a( \overline{c} - \overline{b} ) + \overline{a} ( c-b) = 0 $$

Here I am stuck. Perhaps this is the wrong approach?

Community
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2 Answers2

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Without loss of generality you can assume that $a=1$, because the centre of mass of the three points, the origin, is invariant under their rotation about it. This implies that $c = \overline b$, and that the real component of $b$ and $c$ is $-1/2$.

Théophile
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One possible approach is to write the points in polar form, $a=e^{i\alpha}$, $b=e^{i\beta}$, and $c=e^{i\gamma}$. Now imagine rotating the complex plane about the origin to bring $a$ to $1$. That’s just multiplication by $e^{-i\gamma}$, so the points are now $1,e^{i(\alpha-\gamma)}$, and $e^{i(\beta-\gamma)}$. Moreover, the points still sum to $0$:

$$1+e^{i(\alpha-\gamma)}+e^{i(\beta-\gamma)}=e^{-i\gamma}(a+b+c)=0\;.$$

Now it’s easy to solve for the points $e^{i(\alpha-\gamma)}$ and $e^{i(\beta-\gamma)}$ and verify that they and $1$ are the vertices of an equilateral triangle.

At this point it’s at least intuitively clear that $a,b$, and $c$ must also form an equilateral triangle, but you can finish the job properly by showing that multiplication by $e^{i\theta}$ (for any $\theta$) preserves distances. (This is very easy if you think in polar form.)

Brian M. Scott
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  • Another way of doing pretty much the same is to let $b=\beta a$ and $c=\gamma a$ so that $(1+\beta+\gamma)a=0, a\neq 0$ whence $1+\beta+\gamma=0$. That's just another way of seeing the same thing. It is also then trivial that the imaginary parts of $\beta$ and $\gamma$ have opposite signs and the same magnitude. From there that the real parts are $-\frac 12$ etc if you want a hands-on method. – Mark Bennet Feb 18 '15 at 14:23