Describe the locus of $w$ if $w=\frac{1-z}{1+z}$ and $z=1+iy$ (i.e $z$ is a complex number that moves along the line $x=1$) So I'm trying to solve the following problem: If $z=x+iy$, express $w=\frac{1-z}{1+z}$ in the form $a+bi$ and hence find the equation of the locus of $w$ if $z$ moves along the line $x=1$.
My attempt is:
$w=\frac{1-z}{1+z}\times\frac{1+\bar{z}}{1+\bar{z}}$ 
$=\frac{1-2i\Im{(z)}-|z|^2}{1+2\Re{(z)}+|z|^2}$
$=\frac{1-2yi-x^2-y^2}{1+2x+x^2+y^2}$
$\therefore a+bi=\frac{1-x^2-y^2}{(x+1)^2+y^2}-\frac{2y}{(x+1)^2+y^2}i$
Substituting $x=1$ gives the parametric equations $a=\frac{-y^2}{4+y^2}$ and $b=\frac{-2y}{4+y^2}$
My problem is that I can't reduce these two to only an equation in terms of $a$ and $b$ (best I can do is $b=2ya$, when I try to find $y$ in terms of $b$ it goes $4+y^2=-\frac{2y}{b}$ which is $y^2-\frac{2y}{b}+\left(\frac{y}{b}\right)^2=-4+\left(\frac{y}{b}\right)^2$ and so $\left(y-\frac{y}{b}\right)^2=-4+\left(\frac{y}{b}\right)^2$ but clearly this is not working) so I'm not sure if my steps are right into deriving the parametric equations or if the question just requires me to be more capable of reducing parametric equations. If they can be reduced into equations of just $a$ and $b$, could someone show me how?

$\,w=\frac{1-z}{1+z} \;\;(z \ne -1)\, \iff z=\frac{1-w}{1+w} \;\; (w \ne -1)\,$, then the condition $\,\text{Re}(z) = 1\,$ translates to:
$$
\require{cancel}
2 =  2\text{Re}(z)=z + \bar z = \frac{1-w}{1+w}+\frac{1-\bar w}{1+\bar w}
= \frac{\left(1-|w|^2-\cancel{w}+\bcancel{\bar w}\right)+\left(1-|w|^2+\cancel{w}-\bcancel{\bar w}\right)}{1+|w|^2+w+\bar w}
\\ \iff \cancel{2} + 2|w|^2+2w+2\bar w=\cancel{2} - 2|w|^2 \\ \iff 4 |w|^2+2w+2\bar w=0 \\ \iff (2w+1)(2\bar w+1)=1
$$
The last equation can be written as $\,\left|w+\frac{1}{2}\right|^2=\frac{1}{4}\,$ which is a circle of radius $\,\frac{1}{2}\,$ centered at $\,- \frac{1}{2}\,$.
