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Q)

If $x^{2}+y^{2}=1$ then the value of $\large\frac{1+x+iy}{1+x-iy}$ is

\[\begin{array}{1 1}(1)x-iy&(2)2x\\(3)-2iy&(4)x+iy\end{array}\]

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A)
Given $x^2+y^2=1$
Let $z=x+iy$ the $|z|^2=x^2+y^2=1$
$\large\frac{1}{z}=\frac{1}{x+iy}$
$\qquad= \large\frac{x-iy}{(x+iy)(x-iy)}$
$\qquad= \large\frac{x-iy}{x^2+y^2}$
$\large\frac{1}{z} $$=x-iy=> \large\frac{1}{z}$$=\bar{z}$
$\large\frac{1+x+iy}{1+x-iy}=\frac{1+z}{1+\bar{z}}$
$\qquad= \large\frac{z(1+\bar{z})}{1+ \bar {z}}$
$\qquad= z= x+iy$
Hence 4 is the correct answer.
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