Prove $\forall z\in\mathbb C-\{-1\},\ \left|(z-1)/(z+1)\right|=\sqrt2\iff\left|z+3\right|=\sqrt8$ I'm trying to prove
$$\forall z\in\mathbb C-\{-1\},\ \left|\frac{z-1}{z+1}\right|=\sqrt2\iff\left|z+3\right|=\sqrt8$$
thus showing that the solutions to $\left|(z-1)/(z+1)\right|=\sqrt2$ form the circle of center $-3$ and radius $\sqrt8$. But my memories of algebra in $\mathbb C$ fail me. The simplest I get is writing $z=x+i\,y$ with $(x,y)\in\mathbb R^2-\{(-1,0)\}$ and doing the rather inelegant
$$\begin{align}\left|\frac{z-1}{z+1}\right|=\sqrt2&\iff\left|z-1\right|=\sqrt2\,\left|z+1\right|\\
&\iff\left|z-1\right|^2=2\,\left|z+1\right|^2\\
&\iff(x-1)^2+y^2=2\,((x+1)^2+y^2)\\
&\iff0=x^2+6\,x+y^2+1\\
&\iff(x+3)^2+y^2=8\\
&\iff\left|z+3\right|^2=8\\
&\iff\left|z+3\right|=\sqrt8\\
\end{align}$$
How can I avoid the steps with $x$ and $y$ ?

We have that
$$\frac{z-1}{z+1}=\sqrt 2 e^{i\theta} \implies z=\frac{1+\sqrt 2 e^{i\theta}}{1-\sqrt 2 e^{i\theta}}$$
and $$z+3= \frac{4-2\sqrt 2 e^{i\theta}}{1-\sqrt 2 e^{i\theta}} \implies |z+3|^2=\frac{4-2\sqrt 2 e^{i\theta}}{1-\sqrt 2 e^{i\theta}} \frac{4-2\sqrt 2 e^{-i\theta}}{1-\sqrt 2 e^{-i\theta}} =\frac{24-16\sqrt 2 \cos \theta}{3-2\sqrt 2 \cos \theta}=8$$
