The complex number which satisfies the condition `|(i+z)/(i-z)|=1\ ` lies on a. `c i r c l e\ x^2+y^2=1` b. `t h e\ x-a xi s` c. `t h e\ y-a xi s` d. `t h e\ l in e\ x+y=1`

Options

A. Circle `x^(2) + y^(2) = 1`
B. the X-axis
C. the Y-axis
D. the line `x + y = 1`

Explanation

To solve the problem, we need to analyze the condition given for the complex number \( z \) such that

\[ \left| \frac{i + z}{i - z} \right| = 1. \]

### Step 1: Represent the complex number \( z \)

Let \( z = a + ib \), where \( a \) and \( b \) are real numbers.

### Step 2: Substitute \( z \) into the expression

Substituting \( z \) into the expression gives:

\[ \left| \frac{i + (a + ib)}{i - (a + ib)} \right| = \left| \frac{(a + (1 + b)i)}{(-a + (1 - b)i)} \right|. \]

### Step 3: Simplify the expression

This can be rewritten as:

\[ \left| \frac{a + (1 + b)i}{-a + (1 - b)i} \right| = 1. \]

### Step 4: Cross-multiply the modulus

Using the property of modulus, we can cross-multiply:

\[ |a + (1 + b)i| = |-a + (1 - b)i|. \]

### Step 5: Calculate the moduli

Calculating the moduli gives:

\[ \sqrt{a^2 + (1 + b)^2} = \sqrt{(-a)^2 + (1 - b)^2}. \]

### Step 6: Square both sides

Squaring both sides results in:

\[ a^2 + (1 + b)^2 = a^2 + (1 - b)^2. \]

### Step 7: Simplify the equation

Cancelling \( a^2 \) from both sides gives:

\[ (1 + b)^2 = (1 - b)^2. \]

### Step 8: Expand both sides

Expanding both sides results in:

\[ 1 + 2b + b^2 = 1 - 2b + b^2. \]

### Step 9: Cancel \( b^2 \) and simplify

Cancelling \( b^2 \) from both sides gives:

\[ 1 + 2b = 1 - 2b. \]

### Step 10: Solve for \( b \)

Rearranging gives:

\[ 2b + 2b = 0 \implies 4b = 0 \implies b = 0. \]

### Conclusion

Since \( b = 0 \), this means that the complex number \( z \) lies on the x-axis, where \( z = a + 0i \).

Thus, the answer is:

**b. the x-axis.**