If z(1 + a) = b + ic and a2+b2+c2=1, then (1+iz)(1−iz)=
a+ib1+c
b−ic1+a
a+ic1+b
none of these
1+iz1−iz=1+i(b+ic)/(1+a)1−i(b+ic)/(1+a)=1+a−c+ib1+a+c−ib=(1+a−c+ib)(1+a+c+ib)(1+a+c)2+b2=1+2a+a2−b2−c2+2ib+2iab1+a2+c2+b2+2ac+2(a+c)=2a+2a2+2ib+2iab2+2ac+2(a+c) ∵a2+b2+c2=1=a+a2+ib+iab1+ac+(a+c)=a(a+1)+ib(a+1)(a+1)(c+1)=a+ibc+1