The complex number `z` which satisfies the condition `|(i+z)/(i-z)|=1` lies on the

Options

A. y -axis
B. x-axis
C. line x+y=1
D. circle `x^(2)+y^(2)=1 `

Answer

To solve the problem, we need to find the complex number \( z \) that satisfies the condition \( \left| \frac{i + z}{i - z} \right| = 1 \).

### Step-by-Step Solution:

1. **Understanding the Condition**:
The condition \( \left| \frac{i + z}{i - z} \right| = 1 \) implies that the modulus of the numerator is equal to the modulus of the denominator.

2. **Let \( z = x + iy \)**:
We can express \( z \) in terms of its real part \( x \) and imaginary part \( y \):
\[ z = x + iy \]

3. **Substituting \( z \) into the Condition**:
Substitute \( z \) into the expression:
\[ \left| \frac{i + (x + iy)}{i - (x + iy)} \right| = \left| \frac{(x + i(y + 1))}{(-x + i(1 - y))} \right| \]

4. **Simplifying the Modulus**:
The modulus of a complex number \( \frac{a + bi}{c + di} \) is given by \( \frac{\sqrt{a^2 + b^2}}{\sqrt{c^2 + d^2}} \). Thus, we have:
\[ \left| \frac{i + z}{i - z} \right| = \frac{\sqrt{x^2 + (y + 1)^2}}{\sqrt{x^2 + (1 - y)^2}} \]

5. **Setting Up the Equation**:
Since the moduli are equal, we can set up the equation:
\[ \sqrt{x^2 + (y + 1)^2} = \sqrt{x^2 + (1 - y)^2} \]

6. **Squaring Both Sides**:
Squaring both sides to eliminate the square roots gives:
\[ x^2 + (y + 1)^2 = x^2 + (1 - y)^2 \]

7. **Canceling \( x^2 \)**:
We can cancel \( x^2 \) from both sides:
\[ (y + 1)^2 = (1 - y)^2 \]

8. **Expanding Both Sides**:
Expanding both sides results in:
\[ y^2 + 2y + 1 = 1 - 2y + y^2 \]

9. **Simplifying the Equation**:
Cancel \( y^2 \) and rearranging gives:
\[ 2y + 1 = 1 - 2y \]
\[ 4y = 0 \implies y = 0 \]

10. **Conclusion**:
Since \( y = 0 \), the complex number \( z \) lies on the real axis (the x-axis). Therefore, the solution is:
\[ z = x + 0i \quad \text{(where \( x \) is any real number)} \]

### Final Answer:
The complex number \( z \) lies on the **real axis**.