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

Options

A. `0 lt |z| lt 1`
B. `|z|=1`
C. `|z| gt 1`
D. bounds for `|z|` can not be decided

Answer

To solve the problem, we need to find the condition under which the ratio \(\frac{1-z}{1+z}\) is purely imaginary. Let's denote \(z\) as a complex number \(z = x + iy\), where \(x\) and \(y\) are real numbers.

### Step-by-Step Solution:

1. **Express \(z\) in terms of \(x\) and \(y\)**:
\[ z = x + iy \]

2. **Substitute \(z\) into the ratio**:
\[ \frac{1 - z}{1 + z} = \frac{1 - (x + iy)}{1 + (x + iy)} = \frac{(1 - x) - iy}{(1 + x) + iy} \]

3. **Multiply the numerator and denominator by the conjugate of the denominator**:
\[ \frac{((1 - x) - iy)((1 + x) - iy)}{((1 + x) + iy)((1 + x) - iy)} = \frac{((1 - x)(1 + x) + y^2) + i(-y(1 + x) + y(1 - x))}{(1 + x)^2 + y^2} \]

4. **Simplify the denominator**:
\[ (1 + x)^2 + y^2 = (1 + 2x + x^2 + y^2) \]

5. **Simplify the numerator**:
The real part:
\[ (1 - x)(1 + x) + y^2 = 1 - x^2 + y^2 \]
The imaginary part:
\[ -y(1 + x) + y(1 - x) = -y - xy + y - xy = -2xy \]

6. **Combine the results**:
\[ \frac{(1 - x^2 + y^2) + i(-2xy)}{(1 + 2x + x^2 + y^2)} \]

7. **Set the real part equal to zero for the ratio to be purely imaginary**:
\[ 1 - x^2 + y^2 = 0 \]
Rearranging gives:
\[ x^2 - y^2 = 1 \quad \text{(Equation 1)} \]

8. **Set the imaginary part equal to zero**:
Since we want the imaginary part to be non-zero, we can keep it as:
\[ -2xy \neq 0 \quad \text{(Equation 2)} \]

9. **From Equation 1, we can express \(y^2\) in terms of \(x^2\)**:
\[ y^2 = 1 + x^2 \]

10. **Substituting back into the imaginary part condition**:
We need \(x \neq 0\) and \(y \neq 0\) to satisfy Equation 2.

11. **Conclusion**:
The condition for \(\frac{1-z}{1+z}\) to be purely imaginary is that \(x^2 - y^2 = 1\) and both \(x\) and \(y\) must be non-zero.

### Final Result:
The modulus of \(z\) can be found as:
\[ |z| = \sqrt{x^2 + y^2} = \sqrt{x^2 + (1 + x^2)} = \sqrt{2x^2 + 1} \]
However, the key result is that \(x^2 - y^2 = 1\).