$\displaystyle Consdier\ W=z-\frac{1}{z}+1 \\[2ex] \displaystyle \therefore{}wz+w=z-1 \\[2ex] \displaystyle \therefore{}1+w=z-wz \\[2ex] \displaystyle \therefore{}1+w=z\left(1-w\right) \\[2ex] \displaystyle \\[2ex] \displaystyle \therefore{}z=\frac{1+w}{1-w} \\[2ex] $

$\displaystyle \therefore{}x+iy=1+\frac{(u+iv)}{(1-\left(u+iv\right)}..we\ put\ w=u+iv\ and\ z=x+iy \\[2ex] \displaystyle \therefore{}1-u^2-\frac{v^2}{(1-{u)}^2}+v^2+\frac{i\ v(1+u+1-u)}{(1-{u)}^2}+v^2 \\[2ex] \displaystyle \\[2ex] $

$\displaystyle Comparing\ real\ part\ and\ imaginary\ parts\ on\ both\ the\ sides, \\[2ex] x=1-u^2-\frac{v^2}{(1-{u)}^2}+v^2\ \ and\ y=\ \frac{v(1+u+1-u)}{(1-{u)}^2}+v^2…(1)\\[4ex] $

$Now\ equation\ of\ straight\ line\ is\ y=x\ ...(2) \displaystyle \\[2ex] \displaystyle \therefore{}\ from\ 1\ and\ 2,\ 1-u^2-\frac{v^2}{(1-{u)}^2}+v^2=\ \frac{v(1+u+1-u)}{(1-{u)}^2}+v^2 \\[2ex] \displaystyle \\[2ex] \displaystyle \therefore{}1-u^2-v^2=v\left[2\right] \\[2ex] \displaystyle \\[2ex] \displaystyle \therefore{}0=\left(u^2+v^2+2v-1\right)which\ is\ a\ circle\ in\ the\ w\ plane. \\[2ex] $

$\displaystyle Comparing\ above\ equation\ with\ u^2+v^2+2gu+2fv+c=0,\ we\ get, \\[2ex] \displaystyle 2g=0,2f=2,c=-1 \\[2ex] \displaystyle Centre=(-g,-f)\ =(0,-1) \\[2ex] \displaystyle Radius\ =\ \sqrt{g^2+f^2-c}=\sqrt{0^2+1^2-(-1)}\ =\sqrt{2} \\[2ex] $

$ \displaystyle Hence,\ the\ straight\ line\ y=x\ in\ the\ z\ plane\ is\ mapped\ onto\ the\ circle\ in\ the\ W-plane. \\[2ex] $