# Show that complex numbers are vertices of equilateral triangle

1)Show if $|z_1|=|z_2|=|z_3|=1$ and $z_1+z_2+z_3=0$ then $z_1,z_2,z_3$ are vertices of equilateral triangle inscribed in a circle of radius.

I thought I can take use from roots of unity here, since $|z_1|=|z_2|=|z_3|=1$ they lie at circle at radius $1$ but I don't know how to take advantage from $z_1+z_2+z_3=0$

2)Let $z=\cos\alpha+i\sin\alpha$ where $\alpha \in 0,2\pi$ then find $\arg(z^2-z)$

I come to this siutation $\displaystyle z^2-z=-2\sin{\frac{1}{2}x}(\sin{\frac{3}{2}x}+i\cos{\frac{3}{2}x})=-2\sin{\frac{1}{2}x}(\cos(\frac{\pi}{2}-{\frac{3}{2}x})+i\sin({\frac{\pi}{2}-\frac{3}{2}x}))$ so $\displaystyle 0\le\frac{\pi}{2}-\frac{3}{2}x\le2\pi$ so $\displaystyle\frac{\pi}{3}\ge x \ge - \pi$ so $\displaystyle\arg(z^2-z) =[-\pi,\frac{\pi}{3}]$ ???

• Hint for $1$:You know that $e^{i\alpha},e^{i(\alpha+2\pi 1/3)},e^{i(\alpha+2\pi 2/3)}$ satisfy the requirements. Then assume there are other solutions and find a contradiction. Oct 26, 2014 at 19:30
• Hint for 2: Write z as $z=e^{i\alpha}$ Then $z^2-z = e^{i2\alpha}-e^{i\alpha}$ Oct 26, 2014 at 19:41

Let: $z_1 =e^{ia} ; z_2 = e^{ib}; z_3 = e^{ic}$

$z_1 +z_2 = e^{i\frac{a+b}{2}}*(e^{i\frac{a-b}{2}} + e^{-i\frac{(a-b)}{2}}) = e^{i\frac{a+b}{2}}*2*cos(\frac{a-b}{2}) = -z_3$

=> $|2*cos(\frac{a-b}{2})| = |-z_3| = |z_3| = 1$ ,

If $cos(\frac{a-b}{2}) =\frac{1}{2}$ -> $a = b \pm \frac{2\pi}{3}$ $mod(2\pi)$

here without loss of generality you can assume a= b+ $\frac{2\pi}{3}$ $mod(2\pi)$ (the other case is the same)

you get : $\frac{a+b}{2} = c+\pi$ $mod(2\pi)$ -> b+ $\frac{\pi}{3} = c + \pi$ $mod(2\pi)$ -> $b = c + \frac{2\pi}{3}$ $mod(2\pi)$

You get your equilateral triangle, since you proved that you can rotate of $\frac{2\pi}{3}$ to pass from one point to another. The other cases are exactly the same.

As for 2) , I would use : $z= e^{ia}$

$z^2 - z = e^{2ia} - e^{ia}$ = $e^{\frac{3}{2}ia}*2i*sin(\frac{a}{2})$ = $e^{(\frac{3}{2}a + \frac{\pi}{2})i}*2*sin(\frac{a}{2})$. The sign of the sin is the only thing you have take into account to evaluate correctly the argument. If it is negative, you add $\pi$, else you already have your argument

• why $|\displaystyle 2\cos(\frac{a-b}{2})|=1$ ?? Oct 26, 2014 at 21:57
• You equal the module of both part of the relation : $e^{i\frac{a+b}{2}}*2*cos(\frac{a-b}{2}) = -z_3$ Oct 26, 2014 at 22:46
• I've edited my answer about your second question, if it helps you Oct 27, 2014 at 12:58

Here is a simple way. Let $$z_{1}=\cos\theta_{1}+i\sin\theta_{1}$$,$$z_{2}=\cos\theta_{2}+i\sin\theta_{2}$$, $$z_{3}=\cos\theta_{3}+i\sin\theta_{3}.$$

By $$z_{1}+z_{2}+z_{3}=0$$ we get the sum of the cosines are zero and the sum of the sines is zero.

Squaring the equations

$$\cos\theta_{1}+\cos\theta_{2}=-\cos\theta_{3}$$ and the

$$\sin\theta_{1}+\sin\theta_{2}=-\sin\theta_{3}$$ we get $$\cos(\theta_{1}-\theta_{2})=-\dfrac{1}{2}$$.

Therefore $$\theta_{1}-\theta_{2}=\dfrac{2\pi}{3}$$. Likewise we get similar equations.

Since the triangle with sides $$|z_{1}|$$ and $$|z_{2}|$$ is isosceles, and

the angle $$(z_{1},z_{2})$$ is $$120^{o}$$ then the other two angles are $$30^{o}$$.

We do the same for $$z_{2},z_{3}$$ and we get the same result.

So the angles of the triangle are all $$30^{0}+30^{0}=60^{0}$$ and hence the triangle is equilateral.

Here my extended hint for $2$:

Notice that $z^2-z = z(z-1) = e^{i\alpha}(e^{i\alpha}-1)$. And since $e^{i(\alpha+\beta)} = e^{i\alpha} e^{i\beta}$ therefore $\arg(ab)=\arg(a)+\arg(b)$

So $\arg(z^2-z) = \alpha + \arg(e^{i\alpha}-1)$

You will now find geometrically that $\arg(e^{i\alpha}-1) = \pi/2 + \alpha/2$ (I hope this is correct.)

You just have to consider the triangle $(0,e^{i\alpha},1)$

• OK, I'm getting lost when you say that $\arg(e^{i\alpha}-1) = \pi/2 + \alpha/2$ how do you know that ? Oct 26, 2014 at 20:01
• Make a drawing of the said triangle, you'll notice that it is an isoscles triangle. If you dissect $\alpha$ you get a triangle with a right angle. In this one you can use the sum of the internal angles. Oct 26, 2014 at 20:07
• @flawr If $\alpha=-\pi$ then $Arg(e^{i\alpha}-1)=Arg(-2)=\pi$ and your formula gives $\pi/2 - \pi/2=0$ Oct 26, 2014 at 20:20
• Ok that is an ambiguity but $\alpha$ is assumed to be nonnegative. Oct 26, 2014 at 20:45
• Well same happens if $\alpha=\pi$. Oct 26, 2014 at 21:32

Let:

$$z_{1} = a + ib \\ z_2 =p+iq \\ z_{3} = x + iy$$

Taking the square of modulus:

$$a^ 2 +b^ 2 =p^ 2 +q^ 2 =x^ 2 +y^ 2 =1$$

Equating the real and the imaginary part of $$z_1 + z_2 + z_3$$ to 0,

$$a + p + x = b + q + y = 0$$

$$|z_{1} - z_{2}|^ 2 \\ =(a - p) ^ 2 + (b - q) ^ 2 \\ =2(a^ 2 + p^ 2 )-(a+p)^ 2 + 2(b ^ 2 + q ^ 2) - (b + q) ^ 2 \\ =2(a^ 2 + b^ 2 + p ^ 2 + q ^ 2 ) - x ^ 2 - y ^ 2\\ =4 - 1 =3$$

Similarly, you can show all sides are equal.

A more visual approach to $$1)$$, using roots of unity as OP requested:

If we consider the numbers as vectors in the plane, the conditions $$|z_i|=1$$ and $$z_1+z_2+z_3=0$$ are shown in Figure 1. $$\tag{Fig. 1}$$

Let $$u$$ be any complex number such that $$|u|=1$$, then the transformation of the plane $$z \mapsto uz$$ is a rotation of the plane - and thus also an isometry. This means $$|uz-uw|=|z-w|,$$ for any two complex numbers $$z$$ and $$w$$. In particular, consider the transformation which maps $$z_2$$ to $$\omega=e^{\frac{i2\pi}{3}}$$ as shown in Figure 2. $$\tag{Fig. 2}$$

Since we know the roots of unity form an equilateral triangle $$\triangle_{\omega}$$ in the plane, we need only show $$uz_1$$ and $$uz_3$$ are the other two roots of unity and we're done. To see this, consider the centroid $$G_u$$ of the triangle $$\triangle_u$$ in Figure 2. This is the vector average $$\frac{1}{3}(uz_1+uz_2+uz_3)=\frac{1}{3}u(z_1+z_2+z_3)=0$$ and hence $$G_u=G_{\omega}$$. Now, $$G_u=G_{\omega}$$ is the point $$2/3$$rds of the way from $$uz_2=\omega$$, on the line $$L$$ passing through the midpoints between $$uz_1$$ and $$uz_3$$ and $$\omega^2$$ and $$1$$, respectively - and since only one chord on the unit circle can satisfy these conditions simultaneously, the result follows. $$\square$$