If `z= x+ yi and (|z-1 -i|+4)/(3|z-1-i|-2)=1`, show that `x^(2) + y^(2)- 2x-2y-7=0`

Answer

To solve the problem, we start with the given equation:

\[ \frac{|z - 1 - i| + 4}{3 |z - 1 - i| - 2} = 1 \]

### Step 1: Substitute \( z \)
Let \( z = x + yi \). Then, we can express \( z - 1 - i \) as:

\[ z - 1 - i = (x - 1) + (y - 1)i \]

### Step 2: Calculate the modulus
The modulus of \( z - 1 - i \) is:

\[ |z - 1 - i| = |(x - 1) + (y - 1)i| = \sqrt{(x - 1)^2 + (y - 1)^2} \]

### Step 3: Substitute modulus into the equation
Substituting this into the original equation gives:

\[ \frac{\sqrt{(x - 1)^2 + (y - 1)^2} + 4}{3\sqrt{(x - 1)^2 + (y - 1)^2} - 2} = 1 \]

### Step 4: Cross-multiply
Cross-multiplying yields:

\[ \sqrt{(x - 1)^2 + (y - 1)^2} + 4 = 3\sqrt{(x - 1)^2 + (y - 1)^2} - 2 \]

### Step 5: Rearranging the equation
Rearranging gives:

\[ 4 + 2 = 3\sqrt{(x - 1)^2 + (y - 1)^2} - \sqrt{(x - 1)^2 + (y - 1)^2} \]

This simplifies to:

\[ 6 = 2\sqrt{(x - 1)^2 + (y - 1)^2} \]

### Step 6: Divide by 2
Dividing both sides by 2 gives:

\[ 3 = \sqrt{(x - 1)^2 + (y - 1)^2} \]

### Step 7: Square both sides
Squaring both sides results in:

\[ 9 = (x - 1)^2 + (y - 1)^2 \]

### Step 8: Expand the equation
Expanding the right side gives:

\[ 9 = (x^2 - 2x + 1) + (y^2 - 2y + 1) \]

### Step 9: Combine like terms
Combining like terms results in:

\[ 9 = x^2 + y^2 - 2x - 2y + 2 \]

### Step 10: Rearranging to the standard form
Rearranging gives:

\[ x^2 + y^2 - 2x - 2y - 7 = 0 \]

### Conclusion
Thus, we have shown that:

\[ x^2 + y^2 - 2x - 2y - 7 = 0 \]