If z 1- z 2- z 3 and z 4 be the consecutive vertices of a square- then z 12- z 22- z 32- z 42 equalsA- z1 z2-z2 z3-z3 z4-z4 z1B- z1 z2-z1 z3-z1 z4-z2 z3-z2 z4-z3 z4C- 0D- None of the above

The correct option is A We know that, z_2 z_1/z_4 z_1 = |z_2 z_1|/|z_4 z_1|e^iP/2 = i (as |z_2 z_1| = |z_4 z_1|) ϕ (z_4 z_1)^2 + (z_2 z_1)^2 = 0 similarly z_4 ...

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Byju's Answer

Standard XI

Mathematics

Coordinates of a Point in Space

If z 1, z 2, ...

Question

If $z_{1}, z_{2}, z_{3} a n d z_{4}$ be the consecutive vertices of a square, then $z_{1}^{2} + z_{2}^{2} + z_{3}^{2} + z_{4}^{2}$ equals

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None of the above

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Solution

The correct option is A

We know that, $\frac{z_{2} - z_{1}}{z_{4} - z_{1}}$ = $\frac{| z_{2} - z_{1} |}{| z_{4} - z_{1} |}$ $e^{i P / 2}$ = i (as $| z_{2} - z_{1} |$ = $| z_{4} - z_{1} |)$

$ϕ$ $(z_{4} - z_{1})^{2} + (z_{2} - z_{1})^{2}$ = 0

similarly $\frac{z_{4} - z_{3}}{z_{2} - z_{3}}$ = i

$ϕ$ $(z_{4} - z_{3})^{2} + (z_{2} - z_{3})^{2}$ = 0

On adding Eqs (i) and (ii), we get

$2 z_{1}^{2} + z_{2}^{2} + z_{3}^{2} + z_{4}^{2} - z_{1} z_{2} - z_{4} z_{1} - z_{4} z_{3} - z_{2} z_{3}$ = 0

$z_{1}^{2} + z_{2}^{2} + z_{3}^{2} + z_{4}^{2}$ = $z_{1} z_{2} + z_{2} z_{3} + z_{3} z_{4} + z_{4} z_{1}$

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Similar questions

Q. Let

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

be three complex numbers such that

| z_{1} | = | z_{2} | = | z_{3} | = 1

and

\frac{z_{1}^{2}}{z_{2} z_{3}} + \frac{z_{2}^{2}}{z_{1} z_{3}} + \frac{z_{3}^{2}}{z_{1} z_{2}} = - 1.

Then the possible value(s) of

| z_{1} + z_{2} + z_{3} |

is/are

Q. Let

z_{1}, z_{2}

and

z_{3}

be three complex numbers such that

| z_{1} | = | z_{2} | = | z_{3} | = 1

and

\frac{z_{1}^{2}}{z_{2} z_{3}} + \frac{z_{2}^{2}}{z_{3} z_{1}} + \frac{z_{3}^{2}}{z_{1} z_{2}} + 1 = 0.

Then the sum of all possible values of

| z_{1} + z_{2} + z_{3} |

Q. Let

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

be complex numbers such that

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

and

| z_{1} + z_{2} + z_{3} | = 21.

Given that

| z_{1} - z_{2} | = 2 \sqrt{3}, | z_{1} | = 3 \sqrt{3},

then the value of

| z_{2} |^{2} + | z_{3} |^{2}

Q. Let

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

be three complex numbers such that

| z_{1} | = | z_{2} | = | z_{3} | = 1

and

\frac{z_{1}^{2}}{z_{2} z_{3}} + \frac{z_{2}^{2}}{z_{1} z_{3}} + \frac{z_{3}^{2}}{z_{1} z_{2}} = - 1.

Then the possible value(s) of

| z_{1} + z_{2} + z_{3} |

is/are

Q. Assertion :Let

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

be three complex numbers such that

| 3 z_{1} + 1 | = | 3 z_{2} + 1 | = | 3 z_{3} + 1 |

and

1 + z_{1} + z_{2} + z_{3} = 0

, then

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

will represent vertices of an equilateral triangle on the complex plane. Reason:

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

represent vertices of an equilateral triangle if

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

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If z1,z2,z3 and z4 be the consecutive vertices of a square, then z21+z22+z23+z24 equals

If $z_{1}, z_{2}, z_{3} a n d z_{4}$ be the consecutive vertices of a square, then $z_{1}^{2} + z_{2}^{2} + z_{3}^{2} + z_{4}^{2}$ equals