If one vertex of a square whose diagonals intersect at the origin is 3 cos- - i sin- then other vertices are

Let the vertex A be 3 (cosθ +i sinθ), then OB and OD can be obtained by roatating OA through π2 and π2 respectively. Thus OB = (OA) e^i π/2 and OD=(OA) e^ iπ/ ...

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Standard XIII

Mathematics

Rotation Concept

If one vertex...

Question

If one vertex of a square whose diagonals intersect at the origin is $3 (cos θ + i sin θ)$ , then other vertices are

3 (i sin θ - cos θ)

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3 (i cos θ - sin θ)

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3 (sin θ - i cos θ)

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- 3 (cos θ + i sin θ)

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Solution

The correct option is D $- 3 (cos θ + i sin θ)$

Let the vertex $A$ be $3 (cos θ + i sin θ)$ , then $O B$ and $O D$ can be obtained by roatating $O A$ through $\frac{π}{2}$ and $- \frac{π}{2}$ respectively.
Thus $O B = (O A) e^{i π / 2}$ and $O D = (O A) e^{- i π / 2}$
$\Rightarrow O B = 3 (cos θ + i sin θ) i$ and $O D = 3 (cos θ + i sin θ) (- i)$
$\Rightarrow O B = 3 (- sin θ + i cos θ)$ and $O D = 3 (sin θ - i cos θ)$

Thus, vertices $B$ and $D$ are represented by $\pm 3 (sin θ - i cos θ) .$
Using rotation $O A$ through $π$ we can get $O C$
$∴ O C = (O A) e^{i π} = - O A = - 3 (cos θ + i sin θ)$

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Rotation Concept

Standard XIII Mathematics

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If one vertex of a square whose diagonals intersect at the origin is 3(cosθ+isinθ), then other vertices are

If one vertex of a square whose diagonals intersect at the origin is $3 (cos θ + i sin θ)$ , then other vertices are