If the ratio `(1-z)/(1+z)` is pyrely 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 determine the conditions under which the ratio \(\frac{1 - z}{1 + z}\) is purely imaginary. Let's denote \(z\) as a complex number \(z = x + iy\), where \(x\) is the real part and \(y\) is the imaginary part of \(z\).

### Step-by-Step Solution:

1. **Substituting \(z\)**:
\[ z = x + iy \]
Thus, we have:
\[ 1 - z = 1 - (x + iy) = (1 - x) - iy \]
\[ 1 + z = 1 + (x + iy) = (1 + x) + iy \]

2. **Forming the Ratio**:
The ratio becomes:
\[ \frac{1 - z}{1 + z} = \frac{(1 - x) - iy}{(1 + x) + iy} \]

3. **Multiplying by the Conjugate**:
To simplify the expression, we multiply the numerator and the denominator by the conjugate of the denominator:
\[ \frac{((1 - x) - iy)((1 + x) - iy)}{((1 + x) + iy)((1 + x) - iy)} \]

4. **Calculating the Denominator**:
The denominator simplifies as follows:
\[ (1 + x)^2 + y^2 \]

5. **Calculating the Numerator**:
The numerator expands to:
\[ (1 - x)(1 + x) - (1 - x)iy - iy(1 + x) + y^2 \]
Simplifying gives:
\[ (1 - x^2) + y^2 - i(1 - x + 1 + x)y = (1 - x^2 + y^2) - i(2y) \]

6. **Final Expression**:
Thus, we have:
\[ \frac{(1 - x^2 + y^2) - i(2y)}{(1 + x)^2 + y^2} \]

7. **Condition for Purely Imaginary**:
For the ratio to be purely imaginary, the real part must equal zero:
\[ 1 - x^2 + y^2 = 0 \]
Rearranging gives:
\[ x^2 - y^2 = 1 \]

8. **Finding the Modulus**:
The modulus of \(z\) is given by:
\[ |z| = \sqrt{x^2 + y^2} \]
From the equation \(x^2 - y^2 = 1\), we can express \(x^2\) in terms of \(y^2\):
\[ x^2 = y^2 + 1 \]
Therefore:
\[ |z| = \sqrt{(y^2 + 1) + y^2} = \sqrt{2y^2 + 1} \]

9. **Conclusion**:
The condition \(x^2 + y^2 = 1\) implies that \(z\) lies on the unit circle, hence:
\[ |z| = 1 \]

### Final Answer:
The ratio \(\frac{1 - z}{1 + z}\) is purely imaginary if \( |z| = 1 \).