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AP EAMCET · Maths · Complex Number

If real parts of \(\sqrt{-5-12 i}, \sqrt{5+12 i}\) are positive values, the real part of \(\sqrt{-8-6 i}\) is a negative value and \(a+i b=\frac{\sqrt{-5-12 i}+\sqrt{5+12 i}}{\sqrt{-8-6 i}}\) then \(2 a+b=\)

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

Answer & Solution

Correct Answer

(C) \(-3\)

Step-by-step Solution

Detailed explanation

Let \(\sqrt{-5-12 i}=a+b i, a\gt0\) \(\Rightarrow-5-12 i=(a+b i)^2\) So \(a^2-b^2=-5\) and \(2 a b=-12\) Now, \(a^2+b^2=\sqrt{\left(a^2-b^2\right)^2+4 a^2 b^2}\) \(=\sqrt{(-5)^2+(-12)^2}=\sqrt{169}=13\)…