question_answer

The centre of a regular polygon of \[n\] sides is located at the point \[z=0\] and one of its vertex \[{{z}_{1}}\] is known. If \[{{z}_{2}}\] be the vertex adjacent to \[{{z}_{1}}\], then \[{{z}_{2}}\] is equal to

A) \[{{z}_{1}}\left( \cos \frac{2\pi }{n}\pm i\sin \frac{2\pi }{n} \right)\]

B) \[{{z}_{1}}\left( \cos \frac{\pi }{n}\pm i\sin \frac{\pi }{n} \right)\]

C) \[{{z}_{1}}\left( \cos \frac{\pi }{2n}\pm i\sin \frac{\pi }{2n} \right)\]

D) None of these

Correct Answer: A

Solution :

Let \[A\] be the vertex with affix \[{{z}_{1}}\]. There are two possibilities of \[{{z}_{2}},\]i.e., \[{{z}_{2}}\] can be obtained by rotating \[{{z}_{1}}\] through \[\frac{2\pi }{n}\] either in clockwise or in anticlockwise direction. \ \[\frac{{{z}_{2}}}{{{z}_{1}}}=\left| \frac{{{z}_{2}}}{{{z}_{1}}} \right|\,{{e}^{\pm \frac{i2\pi }{n}}}\] Þ \[{{z}_{2}}={{z}_{1}}{{e}^{\pm \frac{i2\pi }{n}}}\], \[\,(\because \,|{{z}_{2}}|\,=\,|{{z}_{1}}|)\] Þ  \[{{z}_{2}}={{z}_{1}}\left( \cos \frac{2\pi }{n}\pm i\sin \frac{2\pi }{n} \right)\]