Question

Suppose that $z_1, z_2, z_3$ are three vertices of an equilateral triangle in the Argand plane. Let $\alpha=\frac{1}{2}\left(\sqrt{3}+i\right)$ and $\beta$ be a non-zero complex number. The points $\alpha z_{1}+\beta, \alpha z_{2}+\beta, \alpha z_{3}+\beta$ will be

• A. The vertices of an equilateral triangle
• B. The vertices of an isosceles triangle
• C. Collinear
• D. The vertices of an scalene triangle

Answer 1

Answer: (A)

Since, $z_{1}, z_{2}$ and $z_{3}$ are the vertices of an equilateral triangle, therefore
$\left|z_{1}-z_{2}\right| =\left|z_{2}-z_{3}\right| $
$=\left|z_{3}-z_{1}\right|=k $ (say)
Also, $\alpha=\frac{1}{2}(\sqrt{3}+i)$
$\Rightarrow |\alpha| =\frac{1}{2} \sqrt{3+1}=\frac{1}{2} \times 2=1 $
Let $ A =\alpha z_{1}+\beta, B=\alpha z_{2}+\beta$
and $C=\alpha z_{3}+\beta$
Now, $|A B|=\left|\alpha z_{2}+\beta-\left(\alpha z_{1}+\beta\right)\right|$
$=\left|\alpha\left(z_{2}-z_{1}\right)\right|$
$=|\alpha|\left|z_{2}-z_{1}\right|$
$=|1|\left|z_{2}-z_{1}\right|$
$=1\left|z_{2}-z_{1}\right|$
$=\left|z_{2}-z_{1}\right|=k$
Similarly, $B C=C A=k$
Hence, the points $\alpha z_{1}+\beta, \alpha z_{2}+\beta$ and $\alpha z_{3}+\beta$ are the vertices of an equilateral triangle.