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 ! 
 A: Going in steps:


*

*$|\frac12 - z| = \frac c2$, or $|1 - 2z| = c$, is the intuitive representation of a circle.

*$|1 - 2\overline{z}| = c$ is still a circle; we've just reflected it about the real axis.

*Here's the tricky part. Replacing $\overline{z}$ by $\frac1z$ preserves the angle of a point but takes the reciprocal of the magnitude. This makes it an inversion in the complex plane, which also preserves circles. (Well, circles through the origin become lines instead, but our circle isn't one of those.) So $|1 - \frac2z| = c$ is still a circle.

*Finally, going from $|\frac{z-2}{z}| = c$ to $|\frac{z-1}{z+1}| = c$ is just a translation by $1$.

A: 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.
A: Let's have $MA=kMB\iff \dfrac{MA}{MB}=k$ with fixed points $A,B$ and $k>0,k\neq 1$ a constant.
Then the locus of $M$ is a circle.
Note: the case $k=1$ degenerates into a line, indeed $MA=MB$ is simply the perpendicular bisector of $[A,B]$.
The original equation can be written $(\overrightarrow{MA})^2=k^2(\overrightarrow{MB})^2\iff (\overrightarrow{MA}-k\overrightarrow{MB})(\overrightarrow{MA}+k\overrightarrow{MB})=0$

So let's define $I,J$ the centroids $\begin{cases}\overrightarrow{IA}-k\,\overrightarrow{IB}=\vec 0 &:& I=\dfrac{A-kB}{1-k}\\\overrightarrow{JA}+k\,\overrightarrow{JB}=\vec 0&:& J=\dfrac{A+kB}{1+k}\end{cases}$
Then we get $((1-k)\overrightarrow{MI})((1+k)\overrightarrow{MJ})=0\quad$ and since $(1-k^2)\neq 0\ $ (not the degenerate case) then  

$\overrightarrow{MI}\cdot\overrightarrow{MJ}=0\iff$ $M$ is on a circle of diameter $[IJ]$


In the same way $\dfrac{|z-z_A|}{|z-z_B|}=k$ is strictly equivalent to the problem above while identifying $M$ to $z$ and $A,B$ to $z_A,z_B$ respectively.
A: Hint:
$$
\eqalign{
  & \left| {{{z - 1} \over {z + 1}}} \right| = c\quad  \Leftrightarrow \quad \left| {z - 1} \right| = c\left| {z + 1} \right|\quad  \Leftrightarrow   \cr 
  &  \Leftrightarrow \quad {\rm distance}\;\left( {x,y} \right)\;{\rm from}\;(1,0) = c\; \cdot \;{\rm distance}\;\left( {x,y} \right)\;{\rm from}\;( - 1,0) \cr} 
$$
which is another way to define a circle.
Thanks to @Rahul for indicating the actual attribution for such definition ( https://en.wikipedia.org/wiki/Circle#Circle_of_Apollonius )
