Let zk=cos(2kπ10)+isin(2kπ10);k=1,2,⋯,9.List (I)List (II)P. For each zk there exist a zj(1) True such that zk⋅zj=1Q. There exists a k∈{1,2,⋯,9} such that z1⋅z=zk has no solution(2) False z in the set of complex numbers.R.|1−z1||1−z2|⋯|1−z9|10 equals (3)1S.1−9∑k=1cos(2kπ10) equals (4)2Which of the following option is correct?

Let zk=cos(2kπ10)+isin(2kπ10);k=1,2,⋯,9.List (I)List (II)P. For ea

1.	Let zk=cos(2kπ10)+isin(2kπ10);k=1,2,⋯,9.List (I)List (II)P. For each zk there exist a zj(1) True such that zk⋅zj=1Q. There exists a k∈{1,2,⋯,9} such that z1⋅z=zk has no solution(2) False z in the set of complex numbers.R.\|1−z1\|\|1−z2\|⋯\|1−z9\|10 equals (3)1S.1−9∑k=1cos(2kπ10) equals (4)2Which of the following option is correct?
Answer» Let $z_{k} = cos (\frac{2 k π}{10}) + i sin (\frac{2 k π}{10}); k = 1, 2, \dots, 9.$ $\begin{matrix} L i s t (I) & L i s t (I I) P . & For each z_{k} there exist a z_{j} & (1) & True such that z_{k} \cdot z_{j} = 1 Q . & There exists a k \in {1, 2, \dots, 9} such that z_{1} \cdot z = z_{k} has no solution & (2) & False z in the set of complex numbers. R . & \frac{\| 1 - z_{1} \| \| 1 - z_{2} \| \dots \| 1 - z_{9} \|}{10} equals & (3) & 1 S . & 1 - 9 \sum k = 1 cos (\frac{2 k π}{10}) equals & (4) & 2 \end{matrix}$ Which of the following option is correct?