Question

If the line \( \arg(z) = \frac{\pi}{3} \) intersects the curve \( |z - 2\sqrt{3}i| = 2 \), \( z \in \mathbb{C} \), at two distinct points \(A\) and \(B\), then \(AB\) equals:

• A. \(1\)
• B. \(2\)
• C. \(4\)
• D. \(6\)

Hint:

For intersections of a line \( \arg(z)=\theta \) with a circle, parametrize the line using \( z = re^{i\theta} \) to reduce the problem to a quadratic in \(r\).

Answer 1

The Correct Option is B

Concept:
The equation \( \arg(z) = \theta \) represents a straight line through the origin making an angle \( \theta \) with the positive real axis.
The equation \( |z - z_0| = r \) represents a circle with center \( z_0 \) and radius \( r \).
The distance between two intersection points on a straight line can be found using the difference of their parameters.
Step 1: Parametric form of the line Given \( \arg(z) = \frac{\pi}{3} \), any point on the line can be written as: \[ z = re^{i\pi/3} = r\left(\frac{1}{2} + i\frac{\sqrt{3}}{2}\right), \quad r \in \mathbb{R} \] Thus, \[ z = \frac{r}{2} + i\frac{r\sqrt{3}}{2} \]
Step 2: Substitute into the circle equation The circle is: \[ |z - 2\sqrt{3}i| = 2 \] Substitute \( z \): \[ \left| \frac{r}{2} + i\left(\frac{r\sqrt{3}}{2} - 2\sqrt{3}\right) \right| = 2 \]
Step 3: Simplify \[ \left(\frac{r}{2}\right)^2 + \left(\frac{r\sqrt{3}}{2} - 2\sqrt{3}\right)^2 = 4 \] \[ \frac{r^2}{4} + 3\left(\frac{r}{2} - 2\right)^2 = 4 \] \[ \frac{r^2}{4} + \frac{3r^2}{4} - 6r + 12 = 4 \] \[ r^2 - 6r + 8 = 0 \]
Step 4: Solve for \(r\) \[ (r-2)(r-4) = 0 \Rightarrow r = 2,\,4 \]
Step 5: Distance between points The points \(A\) and \(B\) correspond to \( r=2 \) and \( r=4 \). \[ AB = |4 - 2| = 2 \]