Question

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

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

Here is your answer mate.