A) \[4k+3\]
B) \[4k\]
C) \[4k+3\]
D) \[4k+2\]
Correct Answer: B
Solution :
\[z={{1}^{1/n}}={{(\cos 0+i\sin 0)}^{1/n}}\] \[=\cos \frac{2\pi \,r}{n}+i\sin \frac{2\pi r}{n}={{e}^{\frac{i\,2\pi r}{n}}}\] where r varies from 0 to \[n-1\]where each root is unimodular as \[|{{e}^{i\theta }}|=1.\] Let \[{{z}_{1}}=1,{{z}_{2}}={{e}^{i\frac{2\pi \,k}{n}}},\]where \[{{z}_{2}}-0=({{z}_{1}}-0){{e}^{\pi /x}}\] by given condition or \[{{e}^{i\frac{2k\,\pi }{n}}}={{e}^{\frac{\pi }{2}}}\Rightarrow \]\[n=4k\]
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