If - a - ib- - 1 then the simplified form of 1 - b - ai-1 - b - ai is

The correct option is A b + ai Givem |a+i b|=1 √(a^2+b^2)=1 a^2+b^2=1 b^2=1 a^2 Also, 1+b+a i1+b a i=(1+b+a i)(1+b a i)×(1+b ...

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Byju's Answer

Standard XII

Mathematics

Modulus of a Complex Number

If | a + ib...

Question

If $| a + i b | = 1$ then the simplified form of $\frac{1 + b + a i}{1 + b - a i}$ is

b + a i

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a + b i

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(1 + b^{2}) + a^{2}

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a i

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Solution

The correct option is A $b + a i$

$Givem$

$| a + i b | = 1$

$\sqrt{a^{2} + b^{2}} = 1$

$a^{2} + b^{2} = 1$

$b^{2} = 1 - a^{2}$

$Also, \frac{1 + b + a i}{1 + b - a i} = \frac{(1 + b + a i)}{(1 + b - a i)} \times \frac{(1 + b + a i)}{(1 + b + a i)}$

$= \frac{1 + b^{2} + a^{2} (i)^{2} + 2 b + 2 i a + 2 i a b}{(1 + b)^{2} - (a)^{2} (i)^{2}}$

$= \frac{1 + b^{2} + a^{2} + 2 b + 2 i a + 2 i a b}{1^{2} + b^{2} + 2 b + a^{2}}$

$= \frac{2 b^{2} + 2 b + 2 i a (1 + b)}{2 + 2 b}$

$= \frac{2 b (1 + b) + 2 i a (1 + b)}{2 (1 + b)}$

$= b + i a$

option $A$ is correct

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If |a+ib|=1 then the simplified form of 1+b+ai1+b−ai is

The correct option is A b+ai Givem |a+ib|=1√a2+b2=1a2+b2=1b2=1−a2 Also, 1+b+ai1+b−ai=(1+b+ai)(1+b−ai)×(1+b+ai)(1+b+ai)=1+b2+a2(i)2+2b+2ia+2iab(1+b)2−(a)2(i)2=1+b2+a2+2b+2ia+2iab12+b2+2b+a2=2b2+2b+2ia(1+b)2+2b=2b(1+b)+2ia(1+b)2(1+b)=b+iaoption A is correct

If $| a + i b | = 1$ then the simplified form of $\frac{1 + b + a i}{1 + b - a i}$ is