If a2 - b2 - 1- then 1 -b -ia1-b-ia -

The correct option is B b + ia Given a^2 + b^2 = 1 ⇒ #160;1+b+ia/1+b ia =(1+b+ai)^2/(1+b)^2+a^2 = #160;1+b^2 a^2+2b+2iab+2ia/1+b^2 ...

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Definition of Limit

If a2 + b2 ...

Question

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

1

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2

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b + i a

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a + i b

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Solution

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Similar questions

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

\frac{1 + b + i a}{1 + b - i a} =

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

| \to A \times \to B | = \sqrt{3} \to A . \to B,

| \to A + \to B |

{(\frac{a}{b})}^{2} + {(\frac{b}{a})}^{2} - \frac{4}{i} (\frac{a}{b} - \frac{b}{a}) - 6

I_{A}

I_{B}

A & B

A & B

I_{A}

I_{B}

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