Question

A point \(z\) moves in the complex plane such that \(\arg\left(\frac{z-2}{z+2}\right) = \frac{\pi}{4}\), then the minimum value of \(|z - 9\sqrt{2} - 2i|^2\) is equal to _________

Hint:

For any point \(P\) outside a circle with center \(C\) and radius \(R\), the minimum distance to the circumference is \(|CP - R|\).

Answer 1

Correct Answer: 98

Step 1: Understanding the Concept:
The locus \(\arg\left(\frac{z-z_1}{z-z_2}\right) = \theta\) represents an arc of a circle passing through \(z_1\) and \(z_2\).
The expression \(|z - z_0|^2\) represents the square of the distance from the moving point \(z\) to the fixed point \(z_0\).
Step 2: Detailed Explanation:
The points are \(A(-2,0)\) and \(B(2,0)\). Since \(\theta = \pi/4\), the arc is in the upper half-plane.
The center \(C\) lies on the perpendicular bisector of \(AB\), which is the \(y\)-axis (\(x=0\)).
The angle subtended by chord \(AB\) at the center is \(2 \times \frac{\pi}{4} = \frac{\pi}{2}\).
Let the center be \((0, y_0)\). The distance from \((0, y_0)\) to \((2, 0)\) is the radius \(R\).
In \(\triangle OBC\), where \(O\) is origin, \(y_0 = \frac{2}{\tan(\pi/4)} = 2\).
Center \(C = (0, 2)\). Radius \(R = \sqrt{2^2 + 2^2} = 2\sqrt{2}\).
The fixed point is \(P(9\sqrt{2}, 2)\).
The distance from center \(C(0,2)\) to \(P(9\sqrt{2}, 2)\) is:
\[ CP = \sqrt{(9\sqrt{2} - 0)^2 + (2 - 2)^2} = 9\sqrt{2} \]
The minimum distance from point \(P\) to the circle is \(d = CP - R\):
\[ d = 9\sqrt{2} - 2\sqrt{2} = 7\sqrt{2} \]
The minimum value of the square of the distance is:
\[ d^2 = (7\sqrt{2})^2 = 49 \times 2 = 98 \]
Step 3: Final Answer:
The minimum value is 98.