Complex numbers $z_{1}, z_{2}, z_{3}$ are the vertices A, B, C, respectively, of an isosceles right-angled triangle with right angle at C
Then prove that $(z_{1} - z_{2})^{2} = 2 (z_{1} - z_{3}) (z_{3} - z_{2})$ .

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Solution

By coni method
$\frac{z_{2} - z_{3}}{z_{1} - z_{3}} = \frac{B C}{A C} e^{\frac{i π}{2}}$
$\Rightarrow \frac{(z_{2} - z_{3})}{(z_{1} - z_{3})} = i [∵ B C = A C] \Rightarrow (z_{1} - z_{2}) = i (z_{1} - z_{3})$
Squaring both sides
${(z_{2} - z_{3})}^{2} = - {(z_{1} - z_{3})}^{2} \Rightarrow z_{2}^{2} + z_{3}^{2} - {2 z}_{2} z_{3} = - {- z}_{1}^{2} {- z}_{3}^{2} + 2 z_{1} z_{3} \Rightarrow z_{1}^{2} + z_{2}^{2} - {2 z}_{1} z_{2} = {2 z}_{1} z_{3} + {2 z}_{2} z_{3} - {2 z}_{1} z_{2} - 2 z_{3}^{2} {(z_{1} - z_{2})}^{2} = 2 (z_{1} - z_{3}) (z_{3} - z_{2})$

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