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 show that $(z_{1} - z_{2})^{2} = 2 (z_{1} - z_{3}) (z_{3} - z_{2})$ .

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Solution

In right angled triangle ABC
$\frac{z_{2} - z_{3}}{z_{1} - z_{3}} = \frac{B C}{A C} e^{i π / 2}$
$\Rightarrow \frac{(z_{2} - z_{3})}{(z_{1} - z_{3})} = i$ $(∵ B C = A C)$
$\Rightarrow (z_{2} - z_{3}) = i (z_{1} - z_{3})$ squaring
$(z_{2} - z_{3})^{2} = - (z_{1} - z_{3})^{2}$
$\Rightarrow z_{2}^{2} + z_{3}^{2} - 2 z_{2} z_{3} = - z_{1}^{2} - z_{3}^{2} + 2 z_{1} z_{3}$
$\Rightarrow z_{1}^{2} + z_{2}^{2} - 2 z_{1} z_{2} = 2 z_{1} z_{3} + 2 z_{2} z_{3} + 2 z_{2} z_{3} - 2 z_{1} z_{2} - 2 z_{3}^{2}$
$\Rightarrow (z_{1} - z_{2})^{2} = 2 (z_{1} - z_{3}) (z_{3} - z_{2})$
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