If one vertex of a square whose diagonals intersect at the origin is $3\left( {\cos {\rm{\theta }} + i\sin {\rm{\theta }}} \right),$ then the two adjacent vertices are

$ \pm 3\;\left( {\sin {\rm{\theta }} - i\cos {\rm{\theta }}} \right)$

$ \pm \left( {\sin {\rm{\theta }} + i\cos {\rm{\theta }}} \right)$

$ \pm \left( {\cos {\rm{\theta }} - i\sin {\rm{\theta }}} \right)$

JEE MATHEMATICS MCQ: COMPLEX NUMBERS

1 Medium

1 of 1

Application +4 | -1

If one vertex of a square whose diagonals intersect at the origin is

3\left( {\cos \mathrm{\theta } + i\sin \mathrm{\theta }} \right),

then the two adjacent vertices are

View answer and explanation

Correct option: A

\pm 3\;\left( {\sin \mathrm{\theta } - i\cos \mathrm{\theta }} \right)

Option A (

\pm 3\;\left( {\sin \mathrm{\theta } - i\cos \mathrm{\theta }} \right)

) is correct because rotating the given vertex by

\pm \frac{\pi}{2}

gives the adjacent vertices.