<s>[INST] Intuitively, why should I expect a circle in the complex plane from the equation $\left|\frac{z-1}{z+1}\right| = c$? I know how to prove that : ($c \in [0,1[$)
$$C = \{z \in \mathbb{C}: \left|\frac{z-1}{z+1}\right| = c \}$$
is circle in the complex plane. To do so we can for example write $z = x+iy$ and use the brute force approach.
Also, it's worth mentioning that it's intuitive for me that
$$\{z \in \mathbb{C} : \left| z - z_0 \right| = c \}$$
represents a circle. 
But I don't see at all why intuitively $C$ is a circle. So is it possible to understand geometrically why $C$ is a circle in the complex plane ? 
Thank you very much ! 
 [/INST] Let's explore.
$$\bbox{
\left\lvert \frac{z - 1}{z + 1} \right\rvert = c
} \quad \iff \quad \bbox{
\frac{\lvert z - 1 \rvert}{\lvert z + 1 \rvert} = c
} \quad \iff \quad \bbox{
\left\lvert z - 1 \right\rvert = c \left\lvert z + 1 \right\rvert
} \tag{1}\label{NA1}$$
This only makes sense if $0 \lt c \in \mathbb{R}$.
Since $z \in \mathbb{C}$, we can write $z = x + i y$. Since $\lvert z \rvert = \sqrt{x^2 + y^2}$, we have
$$\bbox{
\left\lvert \frac{z - 1}{z + 1} \right\rvert = c
} \quad \iff \quad \bbox{
\sqrt{(x-1)^2 + y^2} = c \sqrt{(x+1)^2 + y^2}
} \tag{2}\label{NA2}$$
Since $x, y, z \in \mathbb{R}$ and $c \gt 0$, both sides are positive, and we can square both sides:
$$\bbox{
\left\lvert \frac{z - 1}{z + 1} \right\rvert = c
} \quad \iff \quad \bbox{
(x-1)^2 + y^2 = c^2 (x+1)^2 + c^2 y^2
} \tag{3}\label{NA3}$$
Expanding and moving all terms to one side, we get
$$\bbox{
\left\lvert \frac{z - 1}{z + 1} \right\rvert = c
} \quad \iff \quad \bbox{
x^2 - 2 x + 1 + y^2 - c^2 x^2 - c^2 y^2 - 2 c^2 x - c^2 = 0
} \tag{4}\label{NA4}$$
The case when $c = 1$ is special, because then $\eqref{NA3}$ simplifies to $x = 0$, which is not a circle but a line (unless you say it is an infinite-radius circle centered at real $\pm\infty$).  In any case, let's continue the exploration with $0 \lt c \in \mathbb{R}$, $c \ne 1$.
We can collect the terms in $\eqref{NA4}$, getting
$$\bbox{
\left\lvert \frac{z - 1}{z + 1} \right\rvert = c
} \quad \iff \quad \bbox{
(1 - c^2)\left( (x + 1)^2 - \frac{4 x}{1 - c^2} + y^2 \right) = 0
} \tag{5}\label{NA5}$$
Because we already decided $c \ne 1$, this is equivalent to
$$\bbox{
\left\lvert \frac{z - 1}{z + 1} \right\rvert = c
} \quad \iff \quad \bbox{
(x + 1)^2 - \frac{4 x}{1 - c^2} + y^2 = 0 , \quad c \ne 1
} \tag{6}\label{NA6}$$
This is getting interesting. Compare to the equation of a circle of radius $r$ centered at $x = x_0$,
$$\bbox{ (x - x_0)^2 + y^2 - r^2 = 0 }$$
Now, if we choose
$$\bbox{ x_0 = \frac{2}{1 - c^2} - 1} , \quad \bbox{ r = \sqrt{\left(\frac{1 + c^2}{1 - c^2} \right)^2 - 1} }$$
we find that
$$\bbox{ (x - x_0)^2 + y^2 - r^2 = (x + 1)^2 - \frac{4 x}{1 - c^2} + y^2 }$$
Therefore,
$$\bbox[#ffffef]{ \bbox{
\left\lvert \frac{z - 1}{z + 1} \right\rvert = c
} \quad \iff \quad \bbox{
(1 - c^2)\left( (x - x_0)^2 + y^2 - r^2 \right) = 0
} , \quad \bbox{
z = x + i y
} \tag{7a}\label{NA7a} }$$
where
$$\bbox[#ffffef]{ \bbox{
x_0 = \frac{2}{1 - c^2} - 1
} , \quad \bbox{
r = \sqrt{\left(\frac{1 + c^2}{1 - c^2} \right)^2 - 1}
} , \quad \bbox{
c \gt 0 
} , \quad \bbox{
c \ne 1
} , \quad \bbox{
c \in \mathbb{R} 
} \tag{7b}\label{NA7b} }$$
and describes a circle of radius $r$ centered at $z = x_0$ on the real axis when $c \gt 0$, $c \ne 1$, and a line along the imaginary axis when $c = 1$.  No intuition or geometry needed, basic algebra suffices.
</s>