Let $α = \cos \frac{2 π}{11} + isin \frac{2 π}{11}, and λ = α^{2019}, μ = α^{2020}, β = α^{2015}$ , then ..(Re(Z)mean real part of complex number Z)

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$2 Re (λ + λ^{2} + λ^{3} + λ^{4} + λ^{5}) + 5 = 3$

$(α - β) (α - β^{2}) \dots \dots (α - β^{10}) + 1 = 1$

$|i + (i - β) (i - β^{2}) \dots (i - β^{10})| = 2$

$(μ - β) (μ - β^{2}) \dots . . (μ - β^{10}) = 1$

answer is B, D.

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Detailed Solution

$\begin{array}{l} A) 1 + α + α^{2} + α^{3} + \dots \dots \dots + α^{10} = 0 \\ \bar{α} = {\bar{α}}^{2} = α^{9} {\bar{α}}^{3} = α^{3}, {\bar{α}}^{4} = α^{7}, {\bar{α}}^{5} = α^{6} \\ so Re (λ + λ^{2} + λ^{3} - λ^{4} + λ^{5}) = Re (α^{6} + α + α^{7} + α^{2} + α^{8}) \\ [(x - α^{2}) (x - α^{4}) (x - α^{6}) (x - α^{8}) (x - α^{10})] \end{array} \begin{array}{l} [(x - α^{12}) (x - α^{14}) (x - α^{15}) (x - α^{18}) (x - α^{20})] \\ = [(x - α^{2}) (x - α^{4}) (x - α^{6}) (x - α^{8}) (x - α^{10})] \\ [(x - α) (x - α^{3}) (x - α^{5}) (x - α^{7}) (x - α^{09})] \\ \frac{α^{77} - 1}{α^{7} - 1} = \prod_{k = 1}^{10} (μ - β^{k}) = 0 \end{array} C) put x = 1 \Rightarrow \frac{(i)^{11} - 1}{i - 1} = \prod_{k = 1}^{10} (i - β^{k}) = \frac{i^{3} - 1}{i - 1} = 1 D) put α = 1 \Rightarrow= \prod_{k = 1}^{10} (α - β^{k}) = \frac{α^{11} - 1}{α - 1} = 0$