Let z k = cos ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ π 10 k 2 + i sin ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ π 10 k 2 ; k = 1, 2, …, 9. List I List II P. For each z k there exists a z j such that z k ⋅ z j = 1 1. True Q. There exists a k ∈ {1, 2, …, 9} such that z 1 ⋅ z = z k has no solution z in the set of complex numbers. 2. False R. 10 z –

Question

Let z k = cos ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ π 10 k 2 + i sin ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ π 10 k 2 ; k = 1, 2, …, 9. List I List II P. For each z k there exists a z j such that z k ⋅ z j = 1 1. True Q. There exists a k ∈ {1, 2, …, 9} such that z 1 ⋅ z = z k has no solution z in the set of complex numbers. 2. False R. 10 z – 1 ..... z – 1 z – 1 9 2 1 equals 3. 1 S. 1 – ∑ = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ π 9 1 k 10 k 2 cos equals 4. 2 P Q R S

Accepted Answer

1 2 3 4

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