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JEE Mains · Maths · STD 11 - 4.1 complex nubers

माना \(z=1+a i, a>0\) एक ऐसी सम्मिश्र संख्या है, कि \(z^{3}\) एक वास्तविक संख्या है, तो योग \(1+z+z^{2}+\ldots . .+z^{11}\) बराबर है

  1. A \(1365\sqrt 3 i\)
  2. B \(-1365\sqrt 3 i\)
  3. C \(-1250\sqrt 3 i\)
  4. D \(1250\sqrt 3 i\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(-1365\sqrt 3 i\)

Step-by-step Solution

Detailed explanation

\(z=1+a i\) \(z^{2}=1-a^{2}+2 a i\) \(z^{2} \cdot z=\left\{\left(1-a^{2}\right)+2 a i\right\}\{1+a i\}\) \(=\left(1-a^{2}\right)+2 a i+\left(1-a^{2}\right) \quad a i-2 a^{2}\) \(\because \quad z^{3}\) is real \(\Rightarrow 2 a+\left(1-a^{2}\right) a=0\)…
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