Question

Plzzz solve the problem​

Answer 1

Given :-

If x + iy = (a + i) / (a - i) , then prove that ay - 1 = x

Solution :-

$\sf x+iy=\dfrac{a+i}{a-i}$

On rationalizing the denominator ,

$:\implies \sf x+iy=\dfrac{a+i}{a-i}\times \dfrac{a+i}{a+i}$

$:\implies \sf x+iy=\dfrac{(a+i)^2}{a^2-i^2}$

$\\ :\implies \sf x+iy=\dfrac{a^2+i^2+2ai}{a^2+1}\ \; [\ \because i^2=-1\ ]$

$\\ :\implies \sf x+iy=\dfrac{a^2-1}{a^2+1}+ \dfrac{2ai}{a^2+1}\ \; [\ \because i^2=-1\ ]$

$\\ :\implies \sf x+iy=\dfrac{a^2-1}{a^2+1}+i\bigg( \dfrac{2a}{a^2+1} \bigg)$

From this , We have ,

$\bullet\ \; \bf \orange{x=\dfrac{a^2-1}{a^2+1}\ \&\ y=\dfrac{2a}{a^2+1}}$

Let's calculate ay - 1 ,

$:\implies \sf a\times \dfrac{2a}{a^2+1}-1$

$:\implies \sf \dfrac{2a^2}{a^2+1}-1$

$:\implies \sf \dfrac{2a^2-(a^2+1)}{a^2+1}$

$:\implies \sf \dfrac{a^2-1}{a^2+1}$

$:\implies \sf x\ \; \bigstar$

So , ay - 1 = x

Hence proved

Answer 2

$\sf{x+iy=\dfrac{a+i}{a-i}}$

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$\sf{x+iy=\Bigg(\dfrac{a+i}{a-i}\Bigg)\Bigg(\dfrac{a+i}{a+i}\Bigg)=\dfrac{{(a+i)}^{2}}{a^2+1}=\dfrac{a^2-1}{a^2+1}+\Bigg(\dfrac{2a}{a^2+1}\Bigg)i}$

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$\sf{\implies{x=\dfrac{a^2-1}{a^2+1}\:and\:y=\dfrac{2a}{a^2+1}}}$

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${\therefore{\sf{\:x+1=\dfrac{a^2-1}{a^2+1}+1=\dfrac{2a^2}{a^2+1}=a\Bigg(\dfrac{2a}{a^2+1}\Bigg)=ay}}}$

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$\therefore\sf{\:ay-1=x}$

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Hence proved..