If `Re ((z + 2i)/(z+4))= 0 ` then z lies on a circle with centre :

Options

A. `(-2, -1)`
B. `(-2,1)`
C. `(2, -1)`
D. `(2,1)`

Answer

To solve the problem, we need to find the center of the circle on which the complex number \( z \) lies, given that the real part of \( \frac{z + 2i}{z + 4} = 0 \).

### Step-by-Step Solution:

1. **Substitute \( z \)**:
Let \( z = x + iy \), where \( x \) and \( y \) are real numbers. Then we can express \( z + 2i \) and \( z + 4 \) as:
\[ z + 2i = x + i(y + 2) \]
\[ z + 4 = (x + 4) + iy \]

2. **Form the Expression**:
Now, we can write the expression:
\[ \frac{z + 2i}{z + 4} = \frac{x + i(y + 2)}{(x + 4) + iy} \]

3. **Multiply by the Conjugate**:
To simplify, multiply the numerator and denominator by the conjugate of the denominator:
\[ \frac{(x + i(y + 2))((x + 4) - iy)}{((x + 4) + iy)((x + 4) - iy)} \]

4. **Calculate the Denominator**:
The denominator simplifies to:
\[ (x + 4)^2 + y^2 \]

5. **Calculate the Numerator**:
The numerator expands to:
\[ (x(x + 4) + 4i(y + 2) - iy(x + 4) + y^2 + 2y) \]
This gives:
\[ (x^2 + 4x + y^2 + 2y) + i(4x + 2y - xy) \]

6. **Separate Real and Imaginary Parts**:
The expression can now be written as:
\[ \frac{x^2 + 4x + y^2 + 2y}{(x + 4)^2 + y^2} + i\frac{4x + 2y - xy}{(x + 4)^2 + y^2} \]

7. **Set the Real Part to Zero**:
We are given that the real part equals zero:
\[ x^2 + 4x + y^2 + 2y = 0 \]

8. **Rearrange the Equation**:
Rearranging gives:
\[ x^2 + 4x + y^2 + 2y + 4 - 4 = 0 \]
\[ (x + 2)^2 + (y + 1)^2 = 5 \]

9. **Identify the Circle**:
This is the equation of a circle with center \( (-2, -1) \) and radius \( \sqrt{5} \).

### Conclusion:
Thus, the center of the circle on which \( z \) lies is \( (-2, -1) \).