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If $\;x+iy = \large\frac{a+ib}{a-ib}\;$, prove that $\;x^{2} + y^{2} =1$

$(a)\;1\qquad(b)\;0\qquad(c)\;2\qquad(d)\;4$

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Answer : $\;1$
Explanation :
We have , $\;x+iy = \large\frac{a+ib}{a-ib}\; $
$= \large\frac{a+ib}{a-ib} \times \large\frac{a+ib}{a+ib}$
$= \large\frac{(a+ib)^{2}}{a^{2} - (-ib)^{2}}$
$ = \large\frac{a^{2} + (ib)^{2} +2iab}{a^{2}+b^{2}}$
$= \large\frac{a^{2}-b^{2}}{a^{2}+b^{2}} + i \large\frac{2ab}{a^{2}+ b^{2}}$
So , that $\; x-iy = \large\frac{a^{2}-b^{2}}{a^{2}+b^{2}} + i \large\frac{2ab}{a^{2}+ b^{2}}$
Therefore ,
$x^{2}+y^{2} = (x+iy)(x-iy)$
$ = \large\frac{(a^{2}-b^{2})^{2}}{(a^{2}+b^{2})^{2}} + i \large\frac{4a^{2}b^{2}}{(a^{2}+ b^{2})^{2}}$
$= \large\frac{(a^{2}+b^{2})^{2}} {(a^{2}+b^{2})^{2}} = 1$
answered Apr 16, 2014 by yamini.v
 

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