First slide

Algebra of complex numbers

Question

The complex numbers $z_{1}, z_{2} a n d z_{3}$ satisfying $\frac{z_{1} - z_{3}}{z_{2} - z_{3}} = \frac{1}{2} (1 - \sqrt{3} i)$ are vertices of a triangle which is

Moderate

A

of area zero
B

right-angled isosceles
C

equilateral
D

obtuse-angle isosceles

Solution

$\begin{aligned} |\frac{z_{1} - z_{3}}{z_{2} - z_{3}}| = |\frac{z_{1} - z_{3}}{z_{2} - z_{3}}| = |\frac{1 - i \sqrt{3}}{2}| \\ = \sqrt{\frac{1}{4} (1 + 3)} = 1 \end{aligned}$
$\Rightarrow |z_{1} - z_{3}| = |z_{2} - z_{3}|$
Also, $\frac{z_{1} - z_{3}}{z_{2} - z_{3}} - 1 = \frac{1 - i \sqrt{3}}{2} - 1$
$\begin{array}{l} \to \frac{z_{1} - z_{2}}{z_{2} - z_{3}} = \frac{- 1 - i \sqrt{3}}{2} \\ \Rightarrow \frac{|z_{1} - z_{2}|}{|z_{2} - z_{3}|} = \sqrt{\frac{1}{4} (1 + 3)} = 1 \\ \Rightarrow |z_{1} - z_{2}| = |z_{2} - z_{3}| \end{array}$
Thus , $|z_{1} - z_{3}| = |z_{2} - z_{3}| = |z_{2} - z_{1}|$
Hence, $z_{1}, z_{2} a n d z_{3}$ are the vertices of an equilateral triangle.

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