If `Im((iz+2)/(z+i))=-1` represents part of a circle with radius r units, then the value of `4r^(2)` is (where, `z in C, z ne i,lm(z)` represents the imaginary part of z and `i^(2)=-1`)

Answer

To solve the problem, we need to analyze the given equation \( \text{Im}\left(\frac{iz + 2}{z + i}\right) = -1 \) and find the radius \( r \) of the circle it represents. We will then compute \( 4r^2 \).

### Step-by-step Solution:

1. **Express \( z \) in terms of its real and imaginary parts**:
Let \( z = x + iy \), where \( x \) and \( y \) are real numbers.

2. **Substitute \( z \) into the equation**:
\[ \frac{iz + 2}{z + i} = \frac{i(x + iy) + 2}{(x + iy) + i} = \frac{ix - y + 2}{x + (y + 1)i} \]

3. **Multiply numerator and denominator by the conjugate of the denominator**:
The conjugate of the denominator \( x + (y + 1)i \) is \( x - (y + 1)i \).
\[ \frac{(ix - y + 2)(x - (y + 1)i)}{(x + (y + 1)i)(x - (y + 1)i)} = \frac{(ix - y + 2)(x - (y + 1)i)}{x^2 + (y + 1)^2} \]

4. **Calculate the denominator**:
\[ x^2 + (y + 1)^2 = x^2 + y^2 + 2y + 1 \]

5. **Expand the numerator**:
\[ (ix - y + 2)(x - (y + 1)i) = ix^2 - ixy - (y^2 + y) + 2x + 2i \]
Collecting real and imaginary parts:
- Real part: \( 2x - y^2 - y \)
- Imaginary part: \( x^2 - xy + 2 \)

6. **Set up the equation for the imaginary part**:
The imaginary part is given by:
\[ \text{Im}\left(\frac{(2x - y^2 - y) + i(x^2 - xy + 2)}{x^2 + y^2 + 2y + 1}\right) = \frac{x^2 - xy + 2}{x^2 + y^2 + 2y + 1} = -1 \]

7. **Cross-multiply to eliminate the fraction**:
\[ x^2 - xy + 2 = - (x^2 + y^2 + 2y + 1) \]
Rearranging gives:
\[ 2x^2 + xy + y^2 + 2y + 3 = 0 \]

8. **Rearranging to standard circle form**:
The equation can be rearranged to:
\[ 2x^2 + y^2 + xy + 2y + 3 = 0 \]
Completing the square will help identify the center and radius.

9. **Finding the center and radius**:
The equation can be compared to the standard form of a circle \( (x - h)^2 + (y - k)^2 = r^2 \). After completing the square, we find the center and radius.

10. **Calculate \( 4r^2 \)**:
If we find \( r^2 = \frac{3}{4} \), then:
\[ 4r^2 = 4 \times \frac{3}{4} = 3 \]

### Final Answer:
Thus, the value of \( 4r^2 \) is \( \boxed{3} \).