Question

Suppose that $z_1,z_2,z_3$ are three vertices of an equilateral triangle in the argand plane. Let $\alpha =\frac{1}{2}(\sqrt{3}+i)$ and $\beta$ be an 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 a scalene triangle

Answer 1

The correct option is A The vertices of an equilateral triangle

Since, $z_1,z_2$ and $z_3$ are the vertices of an equilateral triangle, therefore
$|z_1-z_2|=|z_2-z_3|$
$=|z_3-z_1|=k$ (say)
Also, $\alpha =\frac{1}{2}(\sqrt{3}+i)$
$\Rightarrow \left|\alpha \right|=\frac{1}{2}\sqrt{3+1}=\frac{1}{2}\times 2=1$
Let $A=\alpha z_1+\beta ,B=\alpha z_2+\beta$
and $C=\alpha z_3+\beta$
Now, $\left|AB\right|=|\alpha z_2+\beta -(\alpha z_1+\beta )|$
$=\left|\alpha (z_2-z_1)\right|$
$=\left|\alpha \right||z_2-z_1|$
$=\left|1\right||z_2-z_1|$
$=1|z_2-z_1|$
$=|z_2-z_1|=k$
Similarly, $BC=CA=k$
Hence, the points $\alpha z_1+\beta ,\alpha z_2+\beta$ and $\alpha z_3+\beta$ are the vertices of an equilateral triangle.