1. Key Concepts and Formulas

• Complex Number Representation: $z = x + iy$, where $x = \text{Re}(z)$ and $y = \text{Im}(z)$.
• Modulus of a Complex Number: $|z| = |x+iy| = \sqrt{x^2+y^2}$. Represents the distance from the origin.
• Triangle Inequality: $|z_1 + z_2| \leq |z_1| + |z_2|$. Equality holds when $z_1$ and $z_2$ have the same argument.

2. Step-by-Step Solution

Step 1: Substitute $z = x+iy$ and simplify $\frac{z-2i}{z+2i}$. We substitute $z = x+iy$ into the given expression to separate the real and imaginary parts. This is a standard technique for dealing with conditions on complex numbers. $\frac{z-2i}{z+2i} = \frac{x+iy-2i}{x+iy+2i} = \frac{x+i(y-2)}{x+i(y+2)}$

Step 2: Rationalize the denominator. To find the real part, we multiply the numerator and denominator by the conjugate of the denominator. This eliminates the imaginary term from the denominator. $\frac{x+i(y-2)}{x+i(y+2)} \times \frac{x-i(y+2)}{x-i(y+2)} = \frac{[x+i(y-2)][x-i(y+2)]}{[x+i(y+2)][x-i(y+2)]}$ Expanding the numerator and denominator: $\frac{x^2 -ix(y+2) + ix(y-2) - i^2(y-2)(y+2)}{x^2 - i^2(y+2)^2}$ Since $i^2 = -1$: $\frac{x^2 -ix(y+2) + ix(y-2) + (y-2)(y+2)}{x^2 + (y+2)^2}$ $\frac{x^2 + (y^2-4) + i[-x(y+2) + x(y-2)]}{x^2 + (y+2)^2}$ $\frac{x^2 + y^2 - 4 + i(-xy-2x + xy - 2x)}{x^2 + (y+2)^2} = \frac{x^2 + y^2 - 4 - 4ix}{x^2 + (y+2)^2}$

Step 3: Set the real part equal to zero. We are given that $\text{Re}\left(\frac{z-2i}{z+2i}\right) = 0$. Thus, $\frac{x^2 + y^2 - 4}{x^2 + (y+2)^2} = 0$ This implies $x^2 + y^2 - 4 = 0$, provided that the denominator is not zero.

Step 4: Simplify the equation to find the locus. The equation $x^2 + y^2 - 4 = 0$ simplifies to: $x^2 + y^2 = 4$ This is the equation of a circle centered at the origin $(0,0)$ with radius $2$. So, the locus of $z$ is a circle $|z| = 2$. The denominator $x^2 + (y+2)^2$ cannot be zero, because if it were, $x=0$ and $y=-2$, which implies $z=-2i$. However, substituting $z=-2i$ into the original expression results in division by zero, so $z \neq -2i$.

Step 5: Interpret $|z-(6+8i)|$ geometrically. The expression $|z-(6+8i)|$ represents the distance between the complex number $z$ and the complex number $6+8i$. We want to find the maximum possible value of this distance, given that $z$ lies on the circle $x^2 + y^2 = 4$.

Step 6: Calculate the distance from the origin to $6+8i$. The distance from the origin to the point $(6,8)$ is $|6+8i| = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$

Step 7: Determine the maximum distance. Since the distance from the origin to $6+8i$ is $10$, and the radius of the circle is $2$, the maximum distance between a point on the circle and $6+8i$ is the distance from the origin to $6+8i$ plus the radius of the circle. $\text{Maximum distance} = 10 + 2 = 12$

3. Common Mistakes & Tips

• Forgetting to rationalize: When dealing with complex fractions, always rationalize the denominator to separate the real and imaginary parts.
• Assuming triangle inequality is always tight: The triangle inequality gives an upper bound, but it's important to verify if the maximum is actually achievable.
• Ignoring the denominator: Ensure the denominator is non-zero when setting the real part of a fraction to zero.

4. Summary

We first determined the locus of $z$ to be a circle centered at the origin with radius 2. Then, we found the maximum distance from a point on this circle to the point $6+8i$. This maximum distance is the distance from the origin to $6+8i$ plus the radius of the circle, which equals 12.

5. Final Answer

The final answer is \boxed{12}, which corresponds to option (B).