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AP EAMCET · Maths · Complex Number

If \(z\) is a non-real root of \(x^7=1\), then \(1+3 z+5 z^2+7 z^3+9 z^4+11 z^5+13 z^6=\)

  1. A \(\frac{14}{1-z}\)
  2. B \(\frac{-14}{1-z}\)
  3. C \(\frac{15}{1-z}\)
  4. D \(\frac{-15}{1-z}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{-14}{1-z}\)

Step-by-step Solution

Detailed explanation

\(S = 1+3 z+5 z^2+7 z^3+9 z^4+11 z^5+13 z^6\) \(S = (1+z+z^2+z^3+z^4+z^5+z^6) + (2z+4z^2+6z^3+8z^4+10z^5+12z^6)\) Since \(z^7=1\) and \(z \ne 1\), \(\sum_{k=0}^{6} z^k = \frac{z^7-1}{z-1} = 0\). \(S = 0 + 2(z+2z^2+3z^3+4z^4+5z^5+6z^6)\) Consider the sum…