Given z=q+ir1+p, then p+iq1+r=1+iz1−iz if ( where z is complex number, p≠−1,r≠−1 ,q, r real numbers)

z=q+ir1+P,iz=−r+iq1+p By C.D rule1+iz1−iz=1+p−r+iq1+p+r−iqfrom statp+iq1+r=1+izi−iz=1+p−r+ir1+p+r−iqBy cross multiflicationq(1+p+r)−pq=q(1+r)and p(1+p+r)+q2=(1+r)(1+p−r)⇒p2+q2+r2=1

Given z=q+ir1+p, then p+iq1+r=1+iz1−iz if ( where z is complex number, p≠−1,r≠−1 ,q, r real numbers)

z=q+ir1+P,iz=−r+iq1+p By C.D rule1+iz1−iz=1+p−r+iq1+p+r−iqfrom statp+iq1+r=1+izi−iz=1+p−r+ir1+p+r−iqBy cross multiflicationq(1+p+r)−pq=q(1+r)and p(1+p+r)+q2=(1+r)(1+p−r)⇒p2+q2+r2=1

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Given $z = \frac{q + i r}{1 + p}, t h e n \frac{p + i q}{1 + r} = \frac{1 + i z}{1 - i z}$ if ( where z is complex number, $p \neq - 1, r \neq - 1$ ,q, r real numbers)

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$p^{2} + q^{2} + r^{2} = 2$

$p^{2} + q^{2} - r^{2} = 1$

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

$p^{2} - q^{2} - r^{2} = 2$

answer is A.

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Detailed Solution

$z = \frac{q + i r}{1 + P}, i z = \frac{- r + i q}{1 + p}$

By C.D rule

$\begin{matrix} \frac{1 + i z}{1 - i z} = \frac{1 + p - r + i q}{1 + p + r - i q} \\ f r o m s t a t \\ \frac{p + i q}{1 + r} = \frac{1 + i z}{i - i z} = \frac{1 + p - r + i r}{1 + p + r - i q} \end{matrix}$

By cross multiflication

$\begin{matrix} q (1 + p + r) - p q = q (1 + r) \\ a n d p (1 + p + r) + q^{2} = (1 + r) (1 + p - r) \Rightarrow p^{2} + q^{2} + r^{2} = 1 \end{matrix}$