Question

Suppose that $z_1,z_2,z_3$ are three vertices of an equilateral triangle in the Argand plane.
Let $\alpha =\frac{1}{2}(\sqrt3\,+i)$ abd $\beta$ 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)

$\frac{1}{\left(\alpha\,z_{1}+\beta\right)-\left(\alpha\,z_{2}+\beta\right)}+\frac{1}{\left(\alpha\,z_{3}+\beta\right)-\left(\alpha\,z_{3}+\beta\right)}$
$=\frac{1}{\alpha\left(z_{1}-z_{2}\right)}+\frac{1}{\alpha\left(z_{2}-z_{3}\right)}+\frac{1}{\alpha\left(z_{3}-z_{1}\right)}$
$=\frac{1}{\alpha}\left(\frac{1}{z_{1}-z_{2}}+\frac{1}{z_{2}-z_{3}}+\frac{1}{z_{3}-z_{1}}\right)$
$=\frac{1}{\alpha}\left(0\right)=0$
[$\because z_1, z_2, z_3$ are three vertices of an equilateral triangle]
$\therefore \alpha z_{1}+\beta, \alpha z_{2}+\beta, \alpha z_{3}+\beta$ are vertices of an equilateral triangle