For n1 and n ∈ N , if z_1, z_2, …, z_n are the roots of the equation (z+1)^n=z^n , then ∑_i=1^n fraccot^-1 (2 |Im z_i | )-12 Re z_i=

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For n>1 and n ∈ N , if z_1, z_2, …, z_n are the roots of the equation (z+1)^n=z^n , then ∑_i=1^n fraccot^-1 (2 |Im z_i | )-12 Re z_i=

MARKS Content Team · Accepted Answer

Correct answer: (C) 1/2[π-(π-2) n]. Given, z_1, z_2, z_3, …, z_n are roots of equation (z+1)^n =z^n ⇒ ((z+1)/z )=(1)^1 / n (z+1)/z = (cos (2 k π)/n+i sin (2 k π)/n ) z+1 =Z (cos (2 k π)/n+i sin (2 k π)/n ) z =1/(cos (2 k π)/n+i sin (2 k π)/n-1) z =1/2 (1-i cot (k π)/n ) Re z =1/2, Im(z)= ((-1)/2 cot (k π)/n ) 2 Re(z i) =1, cot^-1[2|Im z i|]=(k π)/n ∑_i=1^v (cot)/(-1)(2(Im z i)-1 2 Re(z i) = (π)/n(1+2+3+… . . n)-n=π ((n)(n+1))/(2(n))-n = π ((n+1))/2-n=1/2[n π+π-2 n]=1/2[π-(π-2) n]

For $n>1$ and $n \in \mathbf{N}$, if $z_1, z_2, \ldots, z_n$ are the roots of the equation $(z+1)^n=z^n$, then $\sum_{i=1}^n \frac{\cot ^{-1}\left(2\left|\operatorname{Im} z_i\right|\right)-1}{2 \operatorname{Re} z_i}=$

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