Vertices and Centroid Relation
Let $Z_1, Z_2, Z_3$ represent the vertices of an equilateral triangle in the complex plane. The centroid is denoted by $Z_0$. The formula for the centroid is:
$ Z_0 = \frac{Z_1 + Z_2 + Z_3}{3} $
From this, we derive:
$ Z_1 + Z_2 + Z_3 = 3Z_0 $
Define $W_k = Z_k - Z_0$ for $k=1, 2, 3$. Summing these differences yields:
$ W_1 + W_2 + W_3 = (Z_1 - Z_0) + (Z_2 - Z_0) + (Z_3 - Z_0) $
$ = (Z_1 + Z_2 + Z_3) - 3Z_0 $
Substituting $Z_1 + Z_2 + Z_3 = 3Z_0$, we get:
$ W_1 + W_2 + W_3 = 3Z_0 - 3Z_0 = 0 $
Equilateral Triangle Properties
For an equilateral triangle, the vectors representing the vertices relative to the centroid ($W_1, W_2, W_3$) are of equal magnitude and spaced $120^\circ$ ($\frac{2\pi}{3}$ radians) apart. Let $W_1$ be one such vector. The others can be expressed using $\omega = e^{i 2\pi/3}$, a complex cube root of unity.
The relations are:
- $W_2 = \omega W_1$
- $W_3 = \omega^2 W_1$
Key properties of $\omega$ are $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.
Sum Calculation using Roots of Unity
The problem asks for the value of $\sum_{k=1}^{3}(Z_k - Z_0)^2$, which is $W_1^2 + W_2^2 + W_3^2$. Substitute the relations for $W_2$ and $W_3$:
$ W_1^2 + W_2^2 + W_3^2 = W_1^2 + (\omega W_1)^2 + (\omega^2 W_1)^2 $
Expand the terms:
$ = W_1^2 + \omega^2 W_1^2 + \omega^4 W_1^2 $
Use the property $\omega^4 = \omega^3 \cdot \omega = 1 \cdot \omega = \omega$. The expression becomes:
$ = W_1^2 (1 + \omega^2 + \omega) $
Apply the property $1 + \omega + \omega^2 = 0$:
$ = W_1^2 (0) $
$ = 0 $
Final Answer Validation
The calculation shows that $\sum_{k=1}^{3}(Z_k - Z_0)^2 = 0$. This corresponds to Option B.
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