Question

Let \(z_1\) and \(z_2\) be two complex numbers such that \( \arg(z_1 - z_2) = \frac{\pi}{4} \) and \(z_1, z_2\) satisfy the equation \(|z-3| = \text{Re}(z)\). Then the imaginary part of \(z_1+z_2\) is equal to _________.

Hint:

The equation \(|z-z_0| = \text{Re}(z)\) or similar forms often define a conic section. Identifying this locus (in this case, a parabola) is the first step. For problems involving a chord connecting two points on a conic, using the difference of the point equations (\(y_1^2 - y_2^2 = \dots\)) is a standard and very effective technique.

Answer 1

Correct Answer: 6

Step 1: Analyze the locus of the complex numbers.
Let \(z = x+iy\). The given equation is \(|z-3| = \text{Re}(z)\). Substituting \(z=x+iy\), we get: \[ |(x-3) + iy| = x \] The modulus on the left is \(\sqrt{(x-3)^2 + y^2}\). \[ \sqrt{(x-3)^2 + y^2} = x \] For the equation to be valid, we must have \(x \ge 0\). Squaring both sides: \[ (x-3)^2 + y^2 = x^2 \] \[ x^2 - 6x + 9 + y^2 = x^2 \] \[ y^2 = 6x - 9 \] This is the equation of a parabola. The points \(z_1\) and \(z_2\) lie on this parabola. Let \(z_1 = x_1 + iy_1\) and \(z_2 = x_2 + iy_2\). Then \(y_1^2 = 6x_1 - 9\) and \(y_2^2 = 6x_2 - 9\). Step 2: Use the argument condition.
We are given that \( \arg(z_1 - z_2) = \frac{\pi}{4} \). The complex number \(z_1 - z_2\) is \((x_1-x_2) + i(y_1-y_2)\). The argument is the angle of the vector connecting \(z_2\) to \(z_1\). \[ \tan(\arg(z_1 - z_2)) = \frac{\text{Im}(z_1 - z_2)}{\text{Re}(z_1 - z_2)} = \frac{y_1 - y_2}{x_1 - x_2} \] Given the argument is \(\pi/4\), we have: \[ \frac{y_1 - y_2}{x_1 - x_2} = \tan(\pi/4) = 1 \] This implies that \(y_1 - y_2 = x_1 - x_2\). This is the slope of the chord connecting the two points on the parabola. Step 3: Combine the parabola equations and the slope information.
We have the two equations for the points on the parabola: 1) \(y_1^2 = 6x_1 - 9\) 2) \(y_2^2 = 6x_2 - 9\) Subtracting equation (2) from (1): \[ y_1^2 - y_2^2 = (6x_1 - 9) - (6x_2 - 9) = 6(x_1 - x_2) \] Factor the left side: \[ (y_1 - y_2)(y_1 + y_2) = 6(x_1 - x_2) \] Since \(z_1 \neq z_2\), \(x_1 - x_2 \neq 0\) and we can substitute \(x_1 - x_2 = y_1 - y_2\): \[ (y_1 - y_2)(y_1 + y_2) = 6(y_1 - y_2) \] Since \(y_1 - y_2 \neq 0\), we can divide both sides by \((y_1 - y_2)\): \[ y_1 + y_2 = 6 \] Step 4: Determine the required value.
We need to find the imaginary part of \(z_1 + z_2\). \[ z_1 + z_2 = (x_1 + x_2) + i(y_1 + y_2) \] \[ \text{Im}(z_1 + z_2) = y_1 + y_2 \] From Step 3, we found \(y_1 + y_2 = 6\).