A) \[\frac{\pi }{2}\]
B) \[\frac{-\pi }{2}\]
C) 0
D) \[\pi \]
Correct Answer: A
Solution :
\[z=-i\sum\limits_{k=1}^{6}{\left( \cos \,\frac{2\pi k}{7}\,+i\sin \,\frac{2\pi k}{7} \right)}\] \[=-i({{\alpha }_{1}}+{{\alpha }_{2}}+{{\alpha }_{3}}\,+{{\alpha }_{4}}\,+{{\alpha }_{4}}\,+{{\alpha }_{5}}\,+{{\alpha }_{6}})\] \[=-i\left( (\underbrace{1+{{\alpha }_{1}}+..............+{{\alpha }_{6}})}_{zero}-1 \right)=i\] \[\therefore \,\,amp\,\,z=\frac{\pi }{2}\]
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