The equation `|((1+i)z-2)/((1+i)z+4)|=k` does not represent a circle when k is

Options

A. 2
B. `pi`
C. e
D. 1

Answer

To determine the value of \( k \) for which the equation

\[ \left| \frac{(1+i)z - 2}{(1+i)z + 4} \right| = k \]

does not represent a circle, we can analyze the equation step by step.

### Step 1: Rewrite the equation
Let \( z = x + iy \) where \( x \) and \( y \) are real numbers. Then, we can express the left-hand side in terms of \( x \) and \( y \):

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

So,

\[ (1+i)z - 2 = (x - y - 2) + i(x + y) \]
\[ (1+i)z + 4 = (x - y + 4) + i(x + y) \]

### Step 2: Calculate the modulus
The modulus of a complex number \( a + bi \) is given by \( \sqrt{a^2 + b^2} \). Therefore, we can find the modulus of the numerator and denominator:

\[ \text{Numerator: } |(x - y - 2) + i(x + y)| = \sqrt{(x - y - 2)^2 + (x + y)^2} \]
\[ \text{Denominator: } |(x - y + 4) + i(x + y)| = \sqrt{(x - y + 4)^2 + (x + y)^2} \]

### Step 3: Set up the equation
Now, we can set up the equation:

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

### Step 4: Square both sides
Squaring both sides gives:

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

### Step 5: Cross-multiply
Cross-multiplying yields:

\[ (x - y - 2)^2 + (x + y)^2 = k^2 \left( (x - y + 4)^2 + (x + y)^2 \right) \]

### Step 6: Expand both sides
Expanding both sides will yield a quadratic equation in \( x \) and \( y \). The left-hand side will be a quadratic form, and the right-hand side will also be a quadratic form multiplied by \( k^2 \).

### Step 7: Identify conditions for a circle
For the equation to represent a circle, the coefficients of \( x^2 \) and \( y^2 \) must be equal and non-zero. If \( k^2 = 1 \), the terms will cancel out, leading to a degenerate case (not a circle).

### Conclusion
Thus, the equation does not represent a circle when \( k = 1 \).

### Final Answer
The value of \( k \) for which the equation does not represent a circle is:

\[ \boxed{1} \]