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

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Solution

Applying rotation about point A, we get,

$\frac{z_{2} - z_{1}}{z_{3} - z_{1}} = \sqrt{2} e^{- i \frac{π}{4}}$

Applying rotation about point B, we get,

$\frac{z_{1} - z_{2}}{z_{3} - z_{2}} = \sqrt{2} e^{i \frac{π}{4}}$

Applying rotation about point C, we get,

$\frac{z_{2} - z_{3}}{z_{1} - z_{3}} = \sqrt{2} e^{i \frac{π}{2}}$

Multiply A and B, we get,

$\frac{z_{2} - z_{1}}{z_{3} - z_{1}} \times \frac{z_{1} - z_{2}}{z_{3} - z_{2}} = \sqrt{2} e^{- i \frac{π}{4}} \times \sqrt{2} e^{i \frac{π}{4}}$

$(z_{1} - z_{2}) (z_{1} - z_{2}) = 2 (z_{1} - z_{3}) (z_{3} - z_{2})$

$∴ (z_{1} - z_{2})^{2} = 2 (z_{1} - z_{3}) (z_{3} - z_{2})$

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