A) 4k + 1
B) 4k + 2
C) 4k + 3
D) 4k
Correct Answer: D
Solution :
\[{{1}^{1/n}}=\cos \frac{2r\pi }{n}+i\sin \frac{2r\pi }{n}\] Let\[{{z}_{1}}=\cos \frac{2{{r}_{1}}\pi }{n}+i\sin \frac{2{{r}_{1}}\pi }{n}\] and \[{{z}_{2}}=\cos \frac{2{{r}_{2}}\pi }{n}+i\sin \frac{2{{r}_{2}}\pi }{n}\]. Then \[\angle \,{{Z}_{1}}O{{Z}_{2}}=amp\,\left( \frac{{{z}_{1}}}{{{z}_{2}}} \right)=amp\,({{z}_{1}})-amp\,({{z}_{2}})\] \[=\frac{2({{r}_{1}}-{{r}_{2}})\pi }{n}=\frac{\pi }{2}\] (Given) \\[n=4({{r}_{1}}-{{r}_{2}})\]=4 × integer, so n is of the form 4 k.
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