JEE Mains · Maths · STD 11 - 4.1 complex nubers
Let the set of all values of \(k \in \mathbb{R}\) such that the equation \(z(\bar{z} + 2 + i) + k(2 + 3i) = 0\), \(z \in \mathbb{C}\), has at least one solution, be the interval \([\alpha, \beta]\). Then \(9(\alpha + \beta)\) is equal to:
- A \(-10\)
- B \(-8\)
- C \(10\sqrt{13}\)
- D \(8\sqrt{13}\)
Answer & Solution
Correct Answer
(A) \(-10\)
Step-by-step Solution
Detailed explanation
Let \(z = x + iy\), then \(\bar{z} = x - iy\). Substituting \(z\) into the given equation: \((x + iy)(x - iy + 2 + i) + k(2 + 3i) = 0\) \(x^2 + y^2 + 2x + ix + 2iy - y + 2k + 3ki = 0\) Separating the real and imaginary parts, we get: Real part: \(x^2 + y^2 + 2x - y + 2k = 0\)…
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