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(A) Let $w = \frac{z-2i}{z+2i}$. Given that $	ext{Re}(w) = 0$,we have $w + \bar{w} = 0$.$\frac{z-2i}{z+2i} + \frac{\bar{z}+2i}{\bar{z}-2i} = 0$$(z-2i

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$12$

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(A) Let $w = \frac{z-2i}{z+2i}$. Given that $	ext{Re}(w) = 0$,we have $w + \bar{w} = 0$.$\frac{z-2i}{z+2i} + \frac{\bar{z}+2i}{\bar{z}-2i} = 0$$(z-2i)(\bar{z}-2i) + (\bar{z}+2i)(z+2i) = 0$$z\bar{z} - 2iz - 2i\bar{z} - 4 + z\bar{z} + 2iz + 2i\bar{z} - 4 = 0$$2|z|^2 - 8 = 0$ $\Rightarrow |z|^2 = 4$ $\Rightarrow |z| = 2$.This represents a circle centered at the origin with radius $r = 2$.We want to find the maximum value of $|z - (6+8i)|$,which is the distance from a point $z$ on the circle to the point $P = 6+8i$.The distance from the origin $O(0,0)$ to $P(6,8)$ is $OP = \sqrt{6^2 + 8^2} = \sqrt{36+64} = 10$.The maximum distance from a point on the circle to $P$ is $OP + r = 10 + 2 = 12$.

Let $z$ be a complex number such that the real part of $\frac{z-2i}{z+2i}$ is zero. Then,the maximum value of $|z-(6+8i)|$ is equal to:

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