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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:

\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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    I guess you haven't yet heard of Möbius transformations? – 2017-01-09
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    @DanielFischer no, I have not. But it looks complicated. This is for a Calculus I class. – 2017-01-09
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    It's not really complicated, but one must get used to a bunch of things to get comfortable with them. Those would give a more general principle from which it would be easy to see what's going on. Well. I suppose complex conjugation and $\lvert z\rvert^2 = z\cdot \overline{z}$ have been treated? Then you can manipulate the inequality $\lvert z-1\rvert^2 < r^2\lvert z + 1\rvert^2$ to get a different description of the set. It's either a disk ($r < 1$), a half-plane ($r = 1$) or the complement of a disk ($r > 1$). – 2017-01-09
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    @DanielFischer Hm, I don't quite see that yet. When I expand $z$ to $a + bi$ I arrive at what resembles a circle formula $\left|z - 1\right|^2 < r^2 \left|z + 1\right|^2 = (a - 1)^2 + b^2 < r^2((a + 1)^2 + b^2)$ but from that the plane disk or complement of a circle are not obvious to me yet. Did you mean to apply some möbius transformation on this equation? – 2017-01-09
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    I meant what dxiv did in their answer. It's usually easier to write things in terms of $z$ and $\overline{z}$ than in terms of real and imaginary part, but it also works in that form. Expand $(a \pm 1)^2$, collect everything to one side, group powers of $a$ and $b$, get (for $r\neq 1$) an inequality $(a-c)^2 + b^2 < \rho^2$ resp. $(a-c)^2 + b^2 > \rho^2$. – 2017-01-10

1 Answers 1

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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.