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I assume the solution to this question might be a more general principle, but the problem itself is very specific and I do not know how to generalize it. Given the following set, there apparently are alternate definitions of the set for various $r$’s, such as $r = 2$, which are to be found.

$$\left\{ z \in \mathbb{C} : \left| \frac{z – 1}{z + 1}\right| < r\right\}$$

However, I do not know how to approach this. Mere algebraic manipulation does not seem to lead anywhere:

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\begin{align*}

\left| \frac{z – 1}{z + 1}\right| & = \left| \frac{(a + bi) – 1}{(a + bi) + 1}\right|\\

& = \left| \frac{(a-1 + bi)}{(a+1 + bi)}\right|\\

& = \left| \frac{((a-1) + bi)((a+1) – bi)}{((a+1) + bi)((a+1) – bi)}\right|\\

& = \left| \frac{(a-1)(a+1) -(a-1)bi + (a+1)bi + b^2 }{(a+1)^2 +b^2}\right|\\

& = \left| \frac{a^2 – 1 -abi + bi + abi + bi + b^2 }{(a+1)^2 +b^2}\right|\\

& = \left| \frac{a^2 – 1 + 2bi + b^2 }{(a+1)^2 +b^2}\right|\\

\end{align*}

What could be a more promising approach to find alternate definitions for the set?

Again, sorry this is such a un-generic question, but maybe there is something generic hidden in the problem.

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**Geometric hint**: $\;|z-1|$ is the distance from $z$ to the point $1 = 1+0 \, i\,$ on the real axis, and $z+1$ is the distance to point $-1 = -1 + 0\,i\,$. The locus of points in the plane with constant ratio $\left|\frac{z-1}{z+1}\right|=r$ between the distances to two fixed points is a circle of Apollonius (including the degenerate case when the ratio is $1$ and the locus is the perpendicular bisector of the segment between the two points). Replacing the equality with an inequality $\left|\frac{z-1}{z+1}\right| \lt r$ will give as locus either the interior or the exterior of the respective circle (or, in the degenerate case, one of the two half-planes).

**Algebraic hint** (elaborating on Daniel Fischer’s comment):

$$

\begin{alignat}{3}

\left| \frac{z – 1}{z + 1}\right| < r & \;\;\iff\;\; |z-1|^2 && \lt r^2 |z+1|^2 \\

& \;\;\iff\;\; (z-1)(\bar z – 1) && \lt r^2(z+1)(\bar z+1) \\

& \;\;\iff\;\; (1-r^2)z \bar z – (1+ r^2)(z+\bar z)+ 1-r^2 && \lt 0

\end{alignat}

$$

If $r=1$ the latter reduces to $z+\bar z \lt 0$ which gives the half-plane $\operatorname{Re}(z) \lt 0$.

Otherwise assume for example $r \lt 1$ so that $\lambda = \frac{1+ r^2}{1-r^2}\,\gt 1\,$ then the above becomes:

$$

\begin{alignat}{3}

z \bar z – \lambda(z+\bar z)+ 1 & \lt 0 \\

(z – \lambda)(\bar z – \lambda) – \lambda^2 + 1 & \lt 0 \\

|z – \lambda|^2 & \lt \lambda^2-1

\end{alignat}

$$

The latter defines the interior of an easily recognizable circle.

The remaining case $r \gt 1$ can be worked out in similar fashion.

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