Question

Prove that
1+iz/1-iz=1-c/a-ib




where a, b, c, ∈ R and z is a complex number such that and a2
+ b2 + c2 = 1, and b + ic = (1 +a)z.​

Answer 1

Step-by-step explanation:

1+iz

=

1−i(b+ic)/(1+a)

1+i(b+ic)/(1+a)

=

1+a+c−ib

1+a−c+ib

=

(1+a+c)

2

+b

2

(1+a−c+ib)(1+a+c+ib)

=

1+a

2

+c

2

+b

2

+2ac+2(a+c)

1+2a+a

2

−b

2

−c

2

+2ib+2iab

=

2+2ac+2(a+c)

2a+2a

2

+2ib+2iab

(∵a

2

+b

2

+c

2

=1)

=

1+ac+(a+c)

a+a

2

+ib+iab

=

(a+1)(c+1)

a(a+1)+ib(a+1)

=

c+1

a+ib