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Square

If z1 , z2 ...

Question

If $z_{1}, z_{2}, z_{3}, z_{4}$ be the vertices of a square in Argand plane, then :

A

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

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B

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

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C

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

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D

z_{2} = (4 - i) z_{1} + (4 + i) z_{3}

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Solution

The correct option is A $z_{2} = (1 - i) z_{1} + (1 + i) z_{3}$
$T i p$ : for such questions, remember that you can visualize the simplest case(as drawn above) and then find the relation this especially works in MCQ type questions.
$∵$ if the relation is always true, then it will definitely be true for the simplest case.
$\frac{z_{1} - z_{2}}{z_{3} - z_{2}} = \frac{2 i y}{- 2 x}$ let $x = y$
$z_{1} - z_{2} = z_{2} i - z_{3} i$
$\Rightarrow z_{1} + z_{3} i = (1 + i) z_{2}$
or $\frac{z_{1} (1 - i) + z_{3} (i + 1)}{2} = z_{2}$ .
Rationalizing.

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Similar questions

z_{1}, z_{2}, z_{3}, z_{4}

be the vertices of a square in Argand plane, then prove that

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

2 z_{4} = (1 - i) z_{1} + (1 + i) z_{3}

z_{1}, z_{2}

z_{3}

are the affixes of three points

A, B

C

respectively and satisfy the condition

| z_{1} - z_{2} | = | z_{1} | + | z_{2} |

| (2 - i) z_{1} + i z_{3} | = | z_{1} | + | (1 - i) z_{1} + i z_{3} |

then prove that

△ A B C

in a right angled.

| z_{2} + i z_{1} | = | z_{1} | + | z_{2} |

| z_{1} | = 3

| z_{2} | = 4

△ A B C

, if affixes of

A, B

C

z_{1}, z_{2}

[\frac{z_{2} - i z_{1}}{1 - i}]

respectively, is,

| z_{2} + i z_{1} | = | z_{1} | + | z_{2} |

| z_{1} | = 3

| z_{2} | = 4

, then the area of

Δ A B C

, if affixes of A, B, and C are

z_{1}, z_{2},

[(z_{2} - i z_{1}) / (1 - i)]

respectively, is

A (z_{1}), B (z_{2})

C (z_{2})

are the three points in argand plane where

| z_{1} + z_{2} | = | z_{1} | - | z_{2} |

| (1 - i) z_{1} + i z_{3} | = | z_{1} | + | z_{3} - z_{1} |,

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