If z be any complex number `( z ne 0)` than arg `(( z- i)/( z + i)) = (pi)/(2)` represents the cure

Options

A. |z| = 1
B. ` | z| = 1, Re (z) gt 0 `
C. ` | z| = 1, Re (z) lt 0 `
D. none of these

Answer

To solve the problem, we need to find the locus of the complex number \( z \) such that

\[ \arg\left(\frac{z - i}{z + i}\right) = \frac{\pi}{2}. \]

### Step 1: Express \( z \) in terms of \( x \) and \( y \)

Let \( z = x + iy \), where \( x \) and \( y \) are real numbers.

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

We substitute \( z \) into the expression:

\[ \arg\left(\frac{(x + iy) - i}{(x + iy) + i}\right) = \arg\left(\frac{x + (y - 1)i}{x + (y + 1)i}\right). \]

### Step 3: Simplify the expression

Now, we can simplify the expression:

\[ \frac{x + (y - 1)i}{x + (y + 1)i}. \]

To eliminate the imaginary unit from the denominator, we multiply the numerator and denominator by the conjugate of the denominator:

\[ \frac{(x + (y - 1)i)(x - (y + 1)i)}{(x + (y + 1)i)(x - (y + 1)i)}. \]

### Step 4: Calculate the denominator

The denominator becomes:

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

### Step 5: Calculate the numerator

The numerator becomes:

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

This simplifies to:

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

### Step 6: Set up the argument condition

Now we have:

\[ \arg\left(\frac{x^2 + (y^2 - 1) + i[2xy]}{x^2 + (y + 1)^2}\right) = \frac{\pi}{2}. \]

This means that the real part of the numerator must be zero:

\[ x^2 + (y^2 - 1) = 0. \]

### Step 7: Solve for \( y \)

From the equation \( x^2 + y^2 - 1 = 0 \), we can rearrange it to get:

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

### Step 8: Identify the locus

The equation \( x^2 + y^2 = 1 \) represents a circle with a radius of 1 centered at the origin (0,0).

### Conclusion

Thus, the locus of the complex number \( z \) such that \( \arg\left(\frac{z - i}{z + i}\right) = \frac{\pi}{2} \) is a circle with radius 1 centered at the origin.