For k=1,2, 9, if we define zk=cos(3k /10)+isin(2k /10), then is it pos – McqOptions

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1.	For k=1,2, 9, if we define zk=cos(3k /10)+isin(2k /10), then is it possible that z1 z=zk has no Solution z?
A.	True
B.	False
Answer» C.

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