Question

The complex numbers $z_1,z_2$ and $z_3$ satisfying $\frac{z_1-z_3}{z_2-z_3}=\frac{1-i\sqrt{3}}{2}$ are the vertices of a triangle which is

• A. of area zero
• B. right angled isosceles
• C. equilateral
• D. obtuse angled isosceles

Answer 1

Answer: (C)

$\frac{z_1-z_3}{z_2-z_3}=\frac{1-i\sqrt{3}}{2}=\frac{(1-i\sqrt{3})(1+i\sqrt{3})}{2(1+i\sqrt{3})}$
$=\frac{1-i^{2}3}{2(1+i\sqrt{3})}=\frac{4}{2(1+i\sqrt{3})}=\frac{2}{(1+i\sqrt{3})}$
$\Rightarrow \frac{z_2-z_3}{z_1-z_3}=\frac{1+i\sqrt{3}}{2}=\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}$
$\Rightarrow \left|\frac{z_2-z_3}{z_1-z_3}\right|=1$ and $\arg \left(\frac{z_2-z_3}{z_1-z_3}\right)=\frac{\pi }{3}$
Hence, the triangle is an equilateral.

Alternate Solution
$\therefore \frac{z_1-z_3}{z_2-z_3}=\frac{1-i\sqrt{3}}{2}$
$\Rightarrow \frac{z_2-z_3}{z_1-z_3}=\frac{2}{1-i\sqrt{3}}=\frac{1+i\sqrt{3}}{2}=\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}$
$\Rightarrow \arg \left(\frac{z_2-z_3}{z_1-z_3}\right)=\frac{\pi }{3}$ and also $\left|\frac{z_2-z_3}{z_1-z_3}\right|=1$
Therefore, triangle is equilateral.