Answer: We have transformation , $w=\dfrac{1}{z+i}$

$\therefore z+i=\dfrac{1}{w}$

$\therefore (x+iy)+i=\dfrac{1}{u+iv}$

$\therefore x+i(y+1)=\dfrac{1}{u+iv}\cdot \dfrac{u-iv}{u+iv}$

$\therefore x+i(y+1)= \dfrac{u-iv}{u^2+v^2}$

$\therefore x=\dfrac{u}{u^2+v^2} \ ,y+1= \dfrac{-v}{u^2+v^2}$

For the real axis i.e. the x-axis , y=0 , 

$\therefore 1= \dfrac{-v}{u^2+v^2}$

$\therefore u^2+v^2+v=0$

$\therefore u^2+v^2+v+\dfrac{1}{4}=\dfrac{1}{4}$

$\therefore u^2+(v+\dfrac{1}{2})^2=\dfrac{1}{4}$

Thus , we can say that the given transformation transforms the real aixs of the z plane into a circle with center at $(0,-\dfrac{1}{2})$ with radius $\dfrac{1}{2}$.