Let Z be a complex number satisfying `| ( z-4)/( z-8)| =1` and `| (z)/( z-2)| = ( 3)/( 2)`
What is `|(z-6)/( z+6)|` equal to ?

Options

A. 3
B. 2
C. 1
D. 0

Answer

To solve the problem, we need to analyze the two given conditions involving the complex number \( z \).

### Step 1: Analyze the first condition
The first condition given is:
\[ \left| \frac{z - 4}{z - 8} \right| = 1 \]
This implies that:
\[ |z - 4| = |z - 8| \]
This means that the point \( z \) is equidistant from the points \( 4 \) and \( 8 \) on the complex plane. The locus of points equidistant from two points is the perpendicular bisector of the segment joining those two points.

The midpoint of \( 4 \) and \( 8 \) is \( 6 \), and the perpendicular bisector is the vertical line \( x = 6 \). Therefore, we can express \( z \) as:
\[ z = 6 + yi \quad \text{(where \( y \) is any real number)} \]

### Step 2: Analyze the second condition
The second condition given is:
\[ \left| \frac{z}{z - 2} \right| = \frac{3}{2} \]
This implies:
\[ |z| = \frac{3}{2} |z - 2| \]
Substituting \( z = 6 + yi \):
\[ |6 + yi| = \frac{3}{2} |(6 + yi) - 2| = \frac{3}{2} |4 + yi| \]
Calculating the moduli, we have:
\[ \sqrt{6^2 + y^2} = \frac{3}{2} \sqrt{4^2 + y^2} \]
Squaring both sides:
\[ 36 + y^2 = \frac{9}{4} (16 + y^2) \]
Expanding the right side:
\[ 36 + y^2 = 36 + \frac{9}{4}y^2 \]
Rearranging gives:
\[ 36 + y^2 - 36 - \frac{9}{4}y^2 = 0 \]
Combining like terms:
\[ y^2 - \frac{9}{4}y^2 = 0 \]
This simplifies to:
\[ -\frac{5}{4}y^2 = 0 \]
Thus, \( y^2 = 0 \) which means \( y = 0 \). Therefore, we find:
\[ z = 6 \]

### Step 3: Calculate \( \left| \frac{z - 6}{z + 6} \right| \)
Now we need to find:
\[ \left| \frac{z - 6}{z + 6} \right| \]
Substituting \( z = 6 \):
\[ \left| \frac{6 - 6}{6 + 6} \right| = \left| \frac{0}{12} \right| = 0 \]

### Final Answer
Thus, the value of \( \left| \frac{z - 6}{z + 6} \right| \) is:
\[ \boxed{0} \]