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

Question

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

Answer 1

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.