If a2+b2=1, then (1+b+i a/1+b-i a)=

Question

If a2+b2=1, then (1+b+i a/1+b-i a)= (A) 1 (B) 2 (C) b+i a (D) a+i b

Tardigrade Admin · Accepted Answer

Given that a2+b2=1, therefore (1+b+i a/1+b-i a)=((1+b+i a)(1+b+i a)/(1+b-i a)(1+b+i a)) =((1+b)2-a2+2 i a(1+b)/1+b2+2 b+a2) =((1-a2)+2 b+b2+2 i a(1+b)/2(1+b)) =(2 b2+2 b+2 i a(1+b)/2(1+b))=b+i a

Q. If $a^{2} + b^{2} = 1$ , then $\frac{1 + b + ia}{1 + b - ia} =$

1

2

$b + ia$

$a + ib$

Q. If a2+b2=1, then 1+b−ia1+b+ia​=

1

2

b+ia

a+ib

Given that a2+b2=1, therefore 1+b−ia1+b+ia​=(1+b−ia)(1+b+ia)(1+b+ia)(1+b+ia)​ =1+b2+2b+a2(1+b)2−a2+2ia(1+b)​ =2(1+b)(1−a2)+2b+b2+2ia(1+b)​ =2(1+b)2b2+2b+2ia(1+b)​=b+ia

Q. If $a^{2} + b^{2} = 1$ , then $\frac{1 + b + ia}{1 + b - ia} =$

$b + ia$

$a + ib$