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Step 1: Represent z in the form x + i y
Let the complex number be written as z = x + i y, where x and y are real numbers.
Step 2: Use the argument condition
We are given
\arg \left(\frac{z - 2}{z + 2}\right) = \frac{\pi}{4}.
This can be rewritten as
\arg(z - 2) - \arg(z + 2) = \frac{\pi}{4}.
Step 3: Convert to tangent form
Writing z - 2 as (x - 2) + i y and z + 2 as (x + 2) + i y, the angle difference becomes:
\tan^{-1}\left(\frac{y}{x - 2}\right) - \tan^{-1}\left(\frac{y}{x + 2}\right) = \frac{\pi}{4}.
Using the formula for the difference of inverse tangents,
\tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1}\left(\frac{a - b}{1 + ab}\right),
we get:
\frac{\left(\frac{y}{x - 2}\right) - \left(\frac{y}{x + 2}\right)}{1 + \left(\frac{y}{x - 2}\right)\left(\frac{y}{x + 2}\right)} = \tan\left(\frac{\pi}{4}\right) = 1.
Step 4: Simplify the equation to find the locus
Simplifying the numerator:
\frac{y}{x-2} - \frac{y}{x+2} = y \cdot \frac{(x+2) - (x-2)}{(x-2)(x+2)} = \frac{4y}{x^2 - 4}.
Then the denominator simplifies as well (including the product of the fractions). The final form leads to:
\frac{4y}{x^2 - 4 + y^2} = 1.
Therefore,
x^2 + y^2 - 4y - 4 = 0.
We can rewrite this as
x^2 + (y^2 - 4y + 4) - 4 = 4,
x^2 + (y - 2)^2 = (2\sqrt{2})^2 = 8.
Hence, the locus is a circle with center (0,\,2) and radius 2\sqrt{2}.
Step 5: Find the minimum value of |z - (9\sqrt{2} + 2i)|^2
The point whose distance from 9\sqrt{2} + 2i we are to minimize is on the circle center (0,2) with radius 2\sqrt{2}.
• The center is C(0,2).
• The point in question is P(9\sqrt{2}, 2).
The distance from C to P is |CP| = 9\sqrt{2}.
Since the radius of the circle is 2\sqrt{2}, the closest point on the circle to P will lie on the line connecting C and P, and the minimal distance from P to any point on the circle is
9\sqrt{2} - 2\sqrt{2} = 7\sqrt{2}.
The question asks for the square of this distance:
\bigl(7\sqrt{2}\bigr)^2 = 49 \cdot 2 = 98.
Final Answer
The minimum value of \left|z - (9\sqrt{2} + 2i)\right|^2 is 98.
