Question

Given $z=\frac{q+ir}{1+p}$ , then $\frac{p+iq}{1+r} =\frac{1+iz}{1-iz}$ if

• 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 1

Answer: (A)

$p^2+q^2+r^2=1$

Answer 2

We have $z=\frac{q+ir}{1+p}$ $\therefore \frac{iz}{1}=\frac{-r+iq}{1+p}$ $\therefore \frac{1+iz}{1-iz}=\frac{1+p-r+iq}{1+p+r-iq}$ $\therefore \frac{p+iq}{1+r}=\frac{1+iz}{1-iz}$ if $\frac{p+iq}{1+r}=\frac{1+p-r+iq}{1+p+r-iq}$ $\Rightarrow p\left(1 + p + r\right) + q^{2} + iq \left(1 + p + r\right) - iqp$ $= \left(1 + r\right) \left(1 + p - r\right) + iq \left(1 + r\right)$ $\Rightarrow p\left(1 + p + r\right) + q^{2}$ $=\left(1+r\right)\left(1+p-r\right)$ and $q(1+ p + r) - qp$ $= q (1 + r)$ [This is clearly true] $\therefore$ the condition is $p(1 + p + r) + q^2$ $= (1 + r) (1 + p - r)$ $\Rightarrow p+p^2 + pr + q^2$ $= 1 + p - r + pr - r^2$ $\Rightarrow p^2+q^2+r^2=1$