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:
- 12
- \(\infty\)
- 10
- 8
The Correct Option is A
Approach Solution - 1
We start with the given condition:
\[ \text{Re}\left( \frac{z - 2i}{z + 2i} \right) = 0 \]
Let \( z = x + iy \). Then,
\[ \frac{x + iy - 2i}{x + iy + 2i} = \frac{x + i(y - 2)}{x + i(y + 2)} \]
Rationalizing the denominator:
\[ \frac{x + i(y - 2)}{x + i(y + 2)} \times \frac{x - i(y + 2)}{x - i(y + 2)} \]
\[ = \frac{x^2 + (y^2 - 4) + i(xy - 2x - xy - 2x)}{x^2 + (y + 2)^2} \]
Now, the real part of the expression is:
\[ \text{Re}\left( \frac{z - 2i}{z + 2i} \right) = \frac{x^2 + y^2 - 4}{x^2 + (y + 2)^2} = 0 \]
This implies:
\[ x^2 + y^2 - 4 = 0 \Rightarrow x^2 + y^2 = 4 \]
Hence, the equation represents a circle with center at the origin and radius 2.
To find the maximum value of \( |z - (6 + 8i)| \):
This represents the maximum distance of the point \( (6, 8) \) from the circle \( x^2 + y^2 = 4 \).
Let the center of the circle be \( O(0, 0) \) and radius \( r = 2 \).
The distance from \( O \) to \( P(6, 8) \) is:
\[ OP = \sqrt{6^2 + 8^2} = 10 \]
Therefore, the maximum distance is:
\[ OP + r = 10 + 2 = 12 \]
Final Answer:
\[ \boxed{12} \]
Approach Solution -2
Given the expression:
\[ \frac{z - 2i}{z + 2i} + \frac{\overline{z} + 2i}{\overline{z} - 2i} = 0, \]
we proceed by simplifying each term. Expanding and multiplying, we obtain:
\[ z\overline{z} - 2i\overline{z} - 2iz + 4(-1) + \overline{z}z + 2zi + 2z\overline{i} + 4(-1) = 0. \]
Combining terms, we get:
\[ 2|z|^2 = 8 \implies |z| = 2. \]
Now, we find the maximum value of \( |z - (6 + 8i)| \):
\[ |z - (6 + 8i)|_{\text{maximum}} = 10 + 2 = 12. \]
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