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(A) Since $z_{1}, z_{2},$ and $z_{3}$ are the vertices of an equilateral triangle, we have $|z_{1} - z_{2}| = |z_{2} - z_{3}| = |z_{3} - z_{1}| = k$.

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the vertices of an equilateral triangle

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(A) Since $z_{1}, z_{2},$ and $z_{3}$ are the vertices of an equilateral triangle, we have $|z_{1} - z_{2}| = |z_{2} - z_{3}| = |z_{3} - z_{1}| = k$. Given $\alpha = \frac{1}{2}(\sqrt{3} + i)$, we find $|\alpha| = \frac{1}{2} \sqrt{(\sqrt{3})^2 + 1^2} = \frac{1}{2} \sqrt{4} = 1$. Let $A = \alpha z_{1} + \beta$, $B = \alpha z_{2} + \beta$, and $C = \alpha z_{3} + \beta$. Then the distance between $A$ and $B$ is $|A - B| = |(\alpha z_{1} + \beta) - (\alpha z_{2} + \beta)| = |\alpha(z_{1} - z_{2})| = |\alpha| |z_{1} - z_{2}| = 1 \cdot k = k$. Similarly, $|B - C| = |\alpha(z_{2} - z_{3})| = k$ and $|C - A| = |\alpha(z_{3} - z_{1})| = k$. Since the distances between the new points are equal, they form an equilateral triangle.

Suppose that $z_{1}, z_{2}, z_{3}$ are three vertices of an equilateral triangle in the Argand plane. Let $\alpha = \frac{1}{2}(\sqrt{3} + i)$ and $\beta$ be a non-zero complex number. The points $\alpha z_{1} + \beta, \alpha z_{2} + \beta, \alpha z_{3} + \beta$ will be

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