Given `z = ( q + ir)/( 1+ p )`, then `( p + iq)/( 1+r ) = ( 1+ iz)/( 1-iz)` , if

Options

A. `p^(2) + q^(2) + r^(2) =1`
B. `p^(2) + q^(2) + r^(2) =2`
C. `p^(2) + q^(2) - r^(2) =1`
D. None of these

Answer

To solve the problem, we need to prove the condition under which the equation holds true. We start with the given expression for \( z \) and manipulate it according to the equation provided.

### Step-by-Step Solution:

1. **Given Expression**:
\[ z = \frac{q + ir}{1 + p} \]

2. **Substituting \( z \) into the Equation**:
We need to show that:
\[ \frac{p + iq}{1 + r} = \frac{1 + iz}{1 - iz} \]
Substitute \( z \) into the right-hand side:
\[ iz = i\left(\frac{q + ir}{1 + p}\right) = \frac{iq - r}{1 + p} \]

3. **Calculating \( 1 + iz \)**:
\[ 1 + iz = 1 + \frac{iq - r}{1 + p} = \frac{(1 + p) + (iq - r)}{1 + p} = \frac{1 + p + iq - r}{1 + p} \]

4. **Calculating \( 1 - iz \)**:
\[ 1 - iz = 1 - \frac{iq - r}{1 + p} = \frac{(1 + p) - (iq - r)}{1 + p} = \frac{1 + p - iq + r}{1 + p} \]

5. **Combining the Results**:
Now we can write:
\[ \frac{1 + iz}{1 - iz} = \frac{\frac{1 + p + iq - r}{1 + p}}{\frac{1 + p - iq + r}{1 + p}} = \frac{1 + p + iq - r}{1 + p - iq + r} \]

6. **Cross Multiplying**:
We need to cross-multiply to simplify:
\[ (p + iq)(1 + p - iq + r) = (1 + p + iq - r)(1 + r) \]

7. **Expanding Both Sides**:
- Left Side:
\[ p(1 + p - iq + r) + iq(1 + p - iq + r) \]
Expanding gives:
\[ p + p^2 + pr - piq + iq + ipq - q^2 - qir \]

- Right Side:
\[ (1 + p + iq - r)(1 + r) = 1 + r + p + pr + iq + iq r - r - r^2 \]
Simplifying gives:
\[ 1 + p + iq + pr - r^2 \]

8. **Equating Both Sides**:
After simplifying both sides, we can equate the real and imaginary parts to find the condition:
\[ p^2 + q^2 + r^2 = 1 \]

### Conclusion:
Thus, the condition that holds true is:
\[ p^2 + q^2 + r^2 = 1 \]