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

Options

A. a) Circle x^(2) + y^(2) =1
B. b) The x axis
C. c) The y-axis
D. d) The line x+y=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 1: Understanding the Condition
The condition \( \left| \frac{i + z}{i - z} \right| = 1 \) implies that the magnitudes of the numerator and denominator are equal. Thus, we can rewrite the condition as:

\[ |i + z| = |i - z|. \]

### Step 2: Express \( z \) in terms of its Real and Imaginary Parts
Let \( z = x + iy \), where \( x \) and \( y \) are real numbers. Substituting \( z \) into the equation gives:

\[ |i + (x + iy)| = |i - (x + iy)|. \]

This simplifies to:

\[ |x + (1 + y)i| = |-x + (1 - y)i|. \]

### Step 3: Calculate the Magnitudes
Now we calculate the magnitudes on both sides:

1. For the left side:

\[ |x + (1 + y)i| = \sqrt{x^2 + (1 + y)^2}. \]

2. For the right side:

\[ |-x + (1 - y)i| = \sqrt{(-x)^2 + (1 - y)^2} = \sqrt{x^2 + (1 - y)^2}. \]

### Step 4: Set the Magnitudes Equal
Setting the two magnitudes equal gives:

\[ \sqrt{x^2 + (1 + y)^2} = \sqrt{x^2 + (1 - y)^2}. \]

### Step 5: Square Both Sides
To eliminate the square roots, we square both sides:

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

### Step 6: Simplify the Equation
Cancel \( x^2 \) from both sides:

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

Expanding both sides results in:

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

### Step 7: Solve for \( y \)
Subtract \( y^2 \) and \( 1 \) from both sides:

\[ 2y = -2y. \]

Adding \( 2y \) to both sides gives:

\[ 4y = 0 \implies y = 0. \]

### Step 8: Conclusion
Since \( y = 0 \), the complex number \( z \) lies on the real axis, which means \( z = x + 0i = x \) where \( x \) is any real number.

Thus, the complex number \( z \) that satisfies the given condition lies on the **x-axis**.

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

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