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If $x^{2}+y^{2}=1$, then what is $\frac{1+x+i y}{1+x-i y}$ equal to?
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Verified Answer
The correct answer is:
$x+i y$
Consider $\frac{1+x+\text { iy }}{1+x-i y}=\frac{(1+x+i y)(1+x+i y)}{(1+x-i y)(1+x+i y)}$
(By Rationalizing)
$=\frac{(1+x)^{2}+i y(1+x)+i y(1+x)-y^{2}}{1+x^{2}+2 x+y^{2}}\left(\because i^{2}=-1\right)$
$=\frac{1+x^{2}+2 x-y^{2}+2 i y(1+x)}{2(1+x)}\left(\because x^{2}+y^{2}=1\right)$
$=\frac{1-y^{2}+2 x+x^{2}+2 i y(1+x)}{2(1+x)}$
$=\frac{2 x^{2}+2 x+2 i y(1+x)}{2(1+x)}=x+i y \quad\left(\because 1-y^{2}=x^{2}\right)$
(By Rationalizing)
$=\frac{(1+x)^{2}+i y(1+x)+i y(1+x)-y^{2}}{1+x^{2}+2 x+y^{2}}\left(\because i^{2}=-1\right)$
$=\frac{1+x^{2}+2 x-y^{2}+2 i y(1+x)}{2(1+x)}\left(\because x^{2}+y^{2}=1\right)$
$=\frac{1-y^{2}+2 x+x^{2}+2 i y(1+x)}{2(1+x)}$
$=\frac{2 x^{2}+2 x+2 i y(1+x)}{2(1+x)}=x+i y \quad\left(\because 1-y^{2}=x^{2}\right)$
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