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JEE Advanced · Mathematics · 3. Complex Numbers

Let \(z\) be a complex number such that the imaginary part of \(z\) is non-zero and \(a=z^{2}+z+1\) is real. Then a cannot take the value

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

Answer & Solution

Correct Answer

(D) \(\frac{3}{4}\)

Step-by-step Solution

Detailed explanation

\(\because \operatorname{Im}(z) \neq 0\) \(\Rightarrow \quad z\) is non-real and equation \(z^{2}+z+(1-a)=0\) will have non-real roots, if \(D < 0\)
\(\Rightarrow 1-4(1-a) < 0\) \(\Rightarrow \quad 4 a < 3 \Rightarrow a < \frac{3}{4}\)
\(\therefore a\) can not take the value \(\frac{3}{4}\).
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