Question

If $z_1, z_2, z_3 \in$ are the vertices of an equilateral triangle, whose centroid is $z_0$, then $\sum_{k=1}^{3}(z_k - z_0)^2$ is equal to

Answer 1

0

Answer 2

To solve this problem, we will use properties of complex numbers related to the geometry of triangles, specifically equilateral triangles.

Let the vertices of the equilateral triangle be $z_1, z_2, z_3$. The centroid $z_0$ of the triangle is given by the formula: $z_0 = \frac{z_1 + z_2 + z_3}{3}$

We need to find the value of the expression $\sum_{k=1}^{3}(z_k - z_0)^2$.

Let's define new complex numbers $Z_k = z_k - z_0$ for $k=1, 2, 3$. The expression we need to evaluate is $Z_1^2 + Z_2^2 + Z_3^2$.

First, let's find the centroid of the triangle formed by $Z_1, Z_2, Z_3$: $\frac{Z_1 + Z_2 + Z_3}{3} = \frac{(z_1 - z_0) + (z_2 - z_0) + (z_3 - z_0)}{3}$ $= \frac{(z_1 + z_2 + z_3) - 3z_0}{3}$ Substitute $z_0 = \frac{z_1 + z_2 + z_3}{3}$ into the expression: $= \frac{(z_1 + z_2 + z_3) - 3\left(\frac{z_1 + z_2 + z_3}{3}\right)}{3}$ $= \frac{(z_1 + z_2 + z_3) - (z_1 + z_2 + z_3)}{3} = \frac{0}{3} = 0$ This means that the triangle with vertices $Z_1, Z_2, Z_3$ is an equilateral triangle whose centroid is at the origin (0,0) in the complex plane.

For an equilateral triangle centered at the origin, its vertices can be represented as $Z_1, Z_2, Z_3$ such that they have the same modulus and their arguments differ by $2\pi/3$. Let $Z_1 = R e^{i\theta}$ for some real $R > 0$ and angle $\theta$. Then the other two vertices can be written as $Z_2 = R e^{i(\theta + 2\pi/3)}$ and $Z_3 = R e^{i(\theta + 4\pi/3)}$. Let $\omega = e^{i2\pi/3}$ be a complex cube root of unity. Then $\omega^2 = e^{i4\pi/3}$ and $\omega^3 = 1$. Also, $1 + \omega + \omega^2 = 0$. So, we can write $Z_2 = Z_1 \omega$ and $Z_3 = Z_1 \omega^2$.

Now, let's calculate the sum $Z_1^2 + Z_2^2 + Z_3^2$: $Z_1^2 + Z_2^2 + Z_3^2 = Z_1^2 + (Z_1\omega)^2 + (Z_1\omega^2)^2$ $= Z_1^2 + Z_1^2\omega^2 + Z_1^2\omega^4$ Since $\omega^4 = \omega^3 \cdot \omega = 1 \cdot \omega = \omega$: $= Z_1^2 + Z_1^2\omega^2 + Z_1^2\omega$ Factor out $Z_1^2$: $= Z_1^2 (1 + \omega^2 + \omega)$ Since $1 + \omega + \omega^2 = 0$: $= Z_1^2 (0) = 0$ Thus, $\sum_{k=1}^{3}(z_k - z_0)^2 = 0$.

Alternatively, we can expand the sum directly: $\sum_{k=1}^{3}(z_k - z_0)^2 = (z_1 - z_0)^2 + (z_2 - z_0)^2 + (z_3 - z_0)^2$ $= (z_1^2 - 2z_1z_0 + z_0^2) + (z_2^2 - 2z_2z_0 + z_0^2) + (z_3^2 - 2z_3z_0 + z_0^2)$ $= (z_1^2 + z_2^2 + z_3^2) - 2z_0(z_1 + z_2 + z_3) + 3z_0^2$ Substitute $z_1 + z_2 + z_3 = 3z_0$: $= (z_1^2 + z_2^2 + z_3^2) - 2z_0(3z_0) + 3z_0^2$ $= (z_1^2 + z_2^2 + z_3^2) - 6z_0^2 + 3z_0^2$ $= (z_1^2 + z_2^2 + z_3^2) - 3z_0^2$ Substitute $z_0 = \frac{z_1 + z_2 + z_3}{3}$: $= (z_1^2 + z_2^2 + z_3^2) - 3\left(\frac{z_1 + z_2 + z_3}{3}\right)^2$ $= (z_1^2 + z_2^2 + z_3^2) - 3\frac{(z_1 + z_2 + z_3)^2}{9}$ $= (z_1^2 + z_2^2 + z_3^2) - \frac{1}{3}(z_1 + z_2 + z_3)^2$ Expand $(z_1 + z_2 + z_3)^2 = z_1^2 + z_2^2 + z_3^2 + 2(z_1z_2 + z_2z_3 + z_3z_1)$: $= (z_1^2 + z_2^2 + z_3^2) - \frac{1}{3}(z_1^2 + z_2^2 + z_3^2 + 2(z_1z_2 + z_2z_3 + z_3z_1))$ $= \frac{3(z_1^2 + z_2^2 + z_3^2) - (z_1^2 + z_2^2 + z_3^2) - 2(z_1z_2 + z_2z_3 + z_3z_1)}{3}$ $= \frac{2(z_1^2 + z_2^2 + z_3^2) - 2(z_1z_2 + z_2z_3 + z_3z_1)}{3}$ $= \frac{2}{3}(z_1^2 + z_2^2 + z_3^2 - z_1z_2 - z_2z_3 - z_3z_1)$ For an equilateral triangle with vertices $z_1, z_2, z_3$, it is a known property that $z_1^2 + z_2^2 + z_3^2 - z_1z_2 - z_2z_3 - z_3z_1 = 0$. Therefore, the expression evaluates to: $\frac{2}{3}(0) = 0$