Question

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

• A. $3(i\sin \theta -\cos \theta )$
• B. $3(i\cos \theta -\sin \theta )$
• C. $3(\sin \theta -i\cos \theta )$
• D. $-3(\cos \theta +i\sin \theta )$

Answer 1

Answer: (B), (C), (D)

The correct options are
B $3(i\cos \theta -\sin \theta )$
C $3(\sin \theta -i\cos \theta )$
D $-3(\cos \theta +i\sin \theta )$


Let the vertex $A$ be $3(\cos \theta +i\sin \theta )$, then $OB$ and $OD$ can be obtained by roatating $OA$ through $\frac{\pi }{2}$ and $-\frac{\pi }{2}$.
Thus $OB=\left(OA\right)e^{i\pi /2}$ and $OD=\left(OA\right)e^{-i\pi /2}$
$\Rightarrow OB=3(\cos \theta +i\sin \theta )i$ and $OD=3(\cos \theta +i\sin \theta )(-i)$
$\Rightarrow OB=3(-\sin \theta +i\cos \theta )$ and $OD=3(\sin \theta -i\cos \theta )$
Thus, vertices $B$ and $D$ are represented by $\pm 3(\sin \theta -i\cos \theta ).$
Using rotation $OA$ through $\pi$ we can get $OC$
$\therefore OC=\left(OA\right)e^{i\pi }=-OA=-3(\cos \theta +i\sin \theta )$