<s>[INST] 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 $y$ ?
 [/INST] Using thrice that $\forall c\in\mathbb C,\, \left|c\right|^2=c\,\bar c$, I got it down to
$$\begin{align}\left|\frac{z-1}{z+1}\right|=\sqrt2&\iff\left|z-1\right|=\sqrt2\,\left|z+1\right|\\
&\iff\left|z-1\right|^2=2\,\left|z+1\right|^2\\
&\iff(z-1)\,(\bar z-1)=2\,(z+1)\,(\bar z+1)\\
&\iff0=z\,\bar z+3\,z+3\,\bar z+1\\
&\iff(z+3)(\bar z+3)=8\\
&\iff\left|z+3\right|^2=8\\
&\iff\left|z+3\right|=\sqrt8\\
\end{align}$$
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