Question
Let $${z_1}{\text{ and }}{z_2}{\text{ be }}{n^{th}}$$ roots of unity which subtend a right angle at the origin. Then $$n$$ must be of the form
A.
$$4k + 1$$
B.
$$4k + 2$$
C.
$$4k +3$$
D.
$$4k$$
Answer :
$$4k$$
Solution :
$$\eqalign{
& {\text{Let }}z = {\left( 1 \right)^{\frac{1}{n}}} = {\left( {\cos 2k\pi + i\sin 2k\pi } \right)^{\frac{1}{n}}} \cr
& \,\,z = \cos \frac{{2k\pi }}{n} + i\sin \frac{{2k\pi }}{n},k = 0,1,2,.....,n - 1. \cr
& Let\,\,\,{z_1} = \cos \left( {\frac{{2{k_1}\pi }}{n}} \right) + i\sin \left( {\frac{{2{k_1}\pi }}{n}} \right) \cr
& {\text{and }}{z_2} = \cos \left( {\frac{{2{k_2}\pi }}{n}} \right) + i\sin \frac{{2{k_2}\pi }}{n} \cr} $$
be the two values of $$z.$$ s.t. they subtend $$\angle $$ of 90° at origin.
$$\eqalign{
& \therefore \,\,\frac{{2{k_1}\pi }}{n} - \frac{{2{k_2}\pi }}{n} = \pm \frac{\pi }{2} \cr
& \Rightarrow \,\,4\left( {{k_1} - {k_2}} \right) = \pm n \cr} $$
As $${k_1}$$ and $${k_2}$$ are integers and $${k_1} \ne {k_2}.$$
$$\therefore \,\,n = 4k,k \in {\text{I}}$$