Question

For k=1,2,…9, if we define zk=cos(3kπ/10)+isin(2kπ/10), then is it possible that z1×z=zk has no Solution z?

(a) True

(b) False

This question was addressed to me during an interview.

This intriguing question comes from Roots of Complex Numbers topic in portion Complex Numbers of Engineering Mathematics

Answer 1

Right option is (b) False

To elaborate: z=zk/z1=e^i(2kπ/10-2π/10)=e^iπ/5(k-1)

For k = 2; z=e^iπ/5, therefore a solution always exists.