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Complex Numbers

If z 1 , z ...

Question

If $z_{1}, z_{2}, z_{3}$ are the vertices of an isosceles triangle and right angled at $z_{2},$ then

A

z_{1}^{2} + z_{3}^{2} + 2 z_{2}^{2} = 2 (z_{1} + z_{3}) z_{2}

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B

z_{1}^{2} + z_{3}^{2} + 2 z_{2}^{2} = 2 (z_{1} + z_{3} - z_{2}) z_{2}

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C

{(z_{1} - z_{2})}^{2} + {(z_{2} - z_{3})}^{2} = 0

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D

\frac{z_{1} - z_{2}}{z_{2} - z_{3}}

is imaginary

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Solution

The correct options are
A $z_{1}^{2} + z_{3}^{2} + 2 z_{2}^{2} = 2 (z_{1} + z_{3}) z_{2}$
B $z_{1}^{2} + z_{3}^{2} + 2 z_{2}^{2} = 2 (z_{1} + z_{3} - z_{2}) z_{2}$
C ${(z_{1} - z_{2})}^{2} + {(z_{2} - z_{3})}^{2} = 0$
D $\frac{z_{1} - z_{2}}{z_{2} - z_{3}}$ is imaginary
Given $A B = B C$ (isosceles triangle), $∠ B = 90^{0}$
$∴ ∠ C = ∠ A = \frac{π}{4}$
$∴ \frac{z_{3} - z_{1}}{z_{2} - z_{1}} = \frac{A C}{A B} e^{i π / 4} = \sqrt{2} e^{i π / 4}$ ...(1)
$\frac{z_{2} - z_{3}}{z_{1} - z_{3}} = \frac{B C}{A C} e^{i π / 4} = \frac{1}{\sqrt{2}} e^{i π / 4}$ ...(2)
From equations (1) and (2), we get $\frac{(\frac{z_{3} - z_{1}}{z_{2} - z_{1}})}{(\frac{z_{2} - z_{3}}{z_{1} - z_{3}})} = \frac{\sqrt{2} e^{i π / 4}}{\frac{1}{\sqrt{2}} e^{i π / 4}} = 2$
$\Rightarrow \frac{z_{3} - z_{1}}{z_{2} - z_{1}} . \frac{z_{1} - z_{3}}{z_{2} - z_{3}} = 2 \Rightarrow - {(z_{3} - z_{1})}^{2} = 2 (z_{2}^{2} - z_{2} z_{3} - z_{1} z_{2} + z_{1} z_{3})$
$\Rightarrow z_{1}^{2} + z_{3}^{2} - 2 z_{1} z_{3} = - 2 z_{2}^{2} + 2 z_{2} z_{3} + 2 z_{1} z_{2} - 2 z_{1} z_{3} \Rightarrow z_{1}^{2} + z_{3}^{2} + 2 z_{2}^{2} = 2 z_{2} (z_{1} + z_{3})$
$\Rightarrow z_{1}^{2} + z_{3}^{2} + 2 z_{2} (z_{1} + z_{3} - z_{2})$
$\Rightarrow z_{1}^{2} + z_{2}^{2} + z_{2}^{2} + z_{3}^{2} - 2 z_{1} z_{2} - 2 z_{2} z_{3} = 0$
$\Rightarrow {(z_{1} - z_{2})}^{2} + {(z_{2} - z_{3})}^{2} = 0$
$\Rightarrow {(\frac{z_{1} - z_{2}}{z_{2} - z_{3}})}^{2} = - 1 = cos π + i π sin π$
$\Rightarrow \frac{z_{1} - z_{2}}{z_{2} - z_{3}} = cos \frac{π}{2} + i sin \frac{π}{2} = i$ (imaginary)
Hence, (a),(b),(c),(d) are correct.

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Similar questions

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