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KVPY2018MathematicsComplex Number

Suppose z is any root of 11 z 8 + 21 i z 7 + 10 i z - 22 = 0 where i = - 1 . Then, S = | z | 2 + | z | + 1 satisfies

Options

  1. AS ≤ 3
  2. B3 < S < 7
  3. C7 ≤ S < 13
  4. DS ≥ 13

Correct answer

B. 3 < S < 7

Step-by-step solution

We have, 11 z 8 + 20 i z 7 + 10 i z - 22 = 0 ⇒ z 8 + 20 11 i z 7 + 10 i 11 z - 2 = 0 Root of equation are z 1 , z 2 , z 3 , z 4 , z 5 , … , z 8 ∑ i = 1 8 z i = 20 11 i z 1 z 2 z 3 z 4 z 5 z 6 z 7 z 8 = 2 ⇒ z 1 z 2 z 3 z 4 z 5 z 6 z 7 z 8 = 2 ⇒ z 1 z 2 z 3 z 4 z 4 z 5 z 6 z 7 z 8 = 2 ∵ 1 < z i < 2 ∴ S = | z | 2 + | z | + 1 ⇒ 1 + 1 + 1 < | z | 2 + | z | + 1 < 2 2 + 2 + 1 ⇒ 3 < | z | 2 + | z | + 1 < 7 ∴ 3 < S < 7

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