A) \[4k+1\]
B) \[4k+2\]
C) \[4k+3\]
D) \[4k\]
Correct Answer: D
Solution :
\[{{\text{n}}^{\text{th}}}\]root of unity\[\cos \frac{2m\pi }{n}+i\sin \frac{2m\pi }{n}\] Let for \[m=k,\] root of unity makes an angle of \[{{90}^{o}}\]with\[\frac{2k\pi }{n}=\frac{\pi }{2}\Rightarrow \]\[n=4k\] For \[m=0\] \[{{z}_{1}}=1\] For \[m=k\] \[{{z}_{2}}\]such that \[{{z}_{2}}O{{z}_{1}}={{90}^{o}}\]You need to login to perform this action.
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