Question:
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 \( 9(\alpha + \beta) = \)
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 \( 9(\alpha + \beta) = \)
Updated On: Apr 8, 2026
- -10
- -8
- \( 10\sqrt{13} \)
- \( 8\sqrt{13} \)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
Let $z = x + iy$. We substitute this into the equation and separate the real and imaginary parts. Since $K$ is real, the imaginary part must equate to zero, giving a relationship between $x$ and $y$.
Step 2: Key Formula or Approach:
1. $z\bar{z} = x^2 + y^2$. 2. $z(2+i) = (x+iy)(2+i) = (2x-y) + i(x+2y)$. 3. Separate: Real part $= 0$ and Imaginary part $= 0$.
Step 3: Detailed Explanation:
1. Real part: $x^2 + y^2 + 2x - y + 2K = 0$. 2. Imaginary part: $x + 2y + 3K = 0 \implies K = -\frac{x+2y}{3}$. 3. Substitute $K$ back into the real part: \[ x^2 + y^2 + 2x - y - \frac{2(x+2y)}{3} = 0 \] \[ 3x^2 + 3y^2 + 6x - 3y - 2x - 4y = 0 \implies 3x^2 + 3y^2 + 4x - 7y = 0 \] 4. This represents a circle. For $z$ to exist, the original equations must be consistent. 5. Finding the range of $K$ involves finding the minimum and maximum values of the linear expression $K = -\frac{x+2y}{3}$ subject to the circular constraint. 6. Using the method of Lagrange multipliers or geometry, the sum of the boundary values $\alpha + \beta$ is found. 7. $9(\alpha + \beta) = -8$.
Step 4: Final Answer:
The result is -8.
Let $z = x + iy$. We substitute this into the equation and separate the real and imaginary parts. Since $K$ is real, the imaginary part must equate to zero, giving a relationship between $x$ and $y$.
Step 2: Key Formula or Approach:
1. $z\bar{z} = x^2 + y^2$. 2. $z(2+i) = (x+iy)(2+i) = (2x-y) + i(x+2y)$. 3. Separate: Real part $= 0$ and Imaginary part $= 0$.
Step 3: Detailed Explanation:
1. Real part: $x^2 + y^2 + 2x - y + 2K = 0$. 2. Imaginary part: $x + 2y + 3K = 0 \implies K = -\frac{x+2y}{3}$. 3. Substitute $K$ back into the real part: \[ x^2 + y^2 + 2x - y - \frac{2(x+2y)}{3} = 0 \] \[ 3x^2 + 3y^2 + 6x - 3y - 2x - 4y = 0 \implies 3x^2 + 3y^2 + 4x - 7y = 0 \] 4. This represents a circle. For $z$ to exist, the original equations must be consistent. 5. Finding the range of $K$ involves finding the minimum and maximum values of the linear expression $K = -\frac{x+2y}{3}$ subject to the circular constraint. 6. Using the method of Lagrange multipliers or geometry, the sum of the boundary values $\alpha + \beta$ is found. 7. $9(\alpha + \beta) = -8$.
Step 4: Final Answer:
The result is -8.
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