If z ( 1 + a ) = b + i c and a 2 + b 2 + c 2 = 1 , then 1 + i z 1 − i z = (where, a , b , c ∈ R and i 2 = - 1 )

Step-by-step solution

1 + i z 1 − i z = 1 + i ( b + i c ) / ( 1 + a ) 1 − i ( b + i c ) / ( 1 + a ) = 1 + a − c + i b 1 + a + c − i b = ( 1 + a − c + i b ) ( 1 + a + c + i b ) ( 1 + a + c ) 2 + b 2 = 1 + 2 a + a 2 − b 2 − c 2 + 2 i b + 2 i a b ) 1 + a 2 + c 2 + b 2 + 2 a c + 2 ( a + c ) = a 2 + b 2 + c 2 + 2 a + a 2 − b 2 − c 2 + 2 i b ( 1 + a ) 1 + 1 + 2 a c + 2 ( a + c ) = 2 a ( a + 1 ) + 2 i b ( 1 + a ) 2 ( 1 + a ) ( 1 + c ) = a + i b 1 + c .