If the number `(z-1)/(z+1)` is purely imaginary, then

Options

A. `|z|=1`
B. `|z| gt 1`
C. `|z| lt 1`
D. `|z| gt 2`

Explanation

To solve the problem, we need to determine the conditions under which the expression \(\frac{z-1}{z+1}\) is purely imaginary. Let's denote \(z\) as a complex number in the form \(z = a + bi\), where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit.

### Step 1: Substitute \(z\) into the expression
We start by substituting \(z = a + bi\) into the expression:

\[ \frac{z-1}{z+1} = \frac{(a + bi) - 1}{(a + bi) + 1} = \frac{(a - 1) + bi}{(a + 1) + bi} \]

### Step 2: Multiply numerator and denominator by the conjugate of the denominator
To simplify the expression, we multiply the numerator and denominator by the conjugate of the denominator \((a + 1) - bi\):

\[ \frac{((a - 1) + bi)((a + 1) - bi)}{((a + 1) + bi)((a + 1) - bi)} \]

### Step 3: Simplify the denominator
The denominator simplifies as follows:

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

### Step 4: Simplify the numerator
Now we simplify the numerator:

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

### Step 5: Combine results
Now we have:

\[ \frac{(a^2 - 1 + b^2) + (2b)i}{(a + 1)^2 + b^2} \]

### Step 6: Set the real part to zero for purely imaginary
For the expression to be purely imaginary, the real part must be zero:

\[ a^2 - 1 + b^2 = 0 \]

### Step 7: Solve for \(a\) and \(b\)
Rearranging gives us:

\[ a^2 + b^2 = 1 \]

This equation represents a circle of radius 1 in the complex plane.

### Final Result
Thus, if \(\frac{z-1}{z+1}\) is purely imaginary, then the complex number \(z\) must satisfy the equation:

\[ a^2 + b^2 = 1 \]