Question

If |a+bi|=1, prove that (1+b+ai)/(1+b-ai)=b+ai​

Answer 1

Given |a+bi|=1

⇒$\sqrt{a^{2}+b^{2} }$ = 1

⇒ a² + b² = 1

To prove,

(1 + b + ai)/(1 + b - ai) = b + ai

Taking L.H.S expression.

Multiplying and dividing with (1 + b + ai)

(1 + b + ai)²/[(1 + b + ai)(1 + b - ai)] = b + ai

[(1 + b)² + (ai)² + 2(1 + b)(ai)]/[(1 + b)² - (ai)²] = b + ai

[1 + b² + 2b - a² + 2ai + 2abi]/[(1 + b² + 2b + a²)] = b + ai

[(1 + b² - a² + 2b) + i(2a + 2ab)]/[(1 + a² + b² + 2b)] = b + ai

Writing 1 as (a² + b²) in numerator and (a² + b²) as 1 in denominator

[(a² + b² + b² - a² + 2b) + i(2a + 2ab)]/[(1 + 1 + 2b)] = b + ai

[(2b² + 2b) + i(2a + 2ab)]/(2 + 2b) = b + ai

[2b(b + 1) + 2ai(b + 1)]/[2(b + 1)] = b + ai

[2(b+1)(b + ai)]/(2(b+1)) = b + ai

Eliminating 2(b + 1) in numerator and denominator.

b + ai = b + ai

Hence proved.