What is the geometric interpretation of the following equation?

Question

What is the geometric interpretation of the following equation?

$\displaystyle\left|\frac{1+z}{1-i\bar{z}}\right|=1$

Answer 1

The equation can be transformed to $$ |1 + z| = |1 -i \bar{z}| \quad (*) $$

which implies for the squares of the lengths: $$ |1 + z|^2 = |1 -i \bar{z}|^2 \iff \\ (1+z)(1+\bar{z}) = (1-i\bar{z})(1+i z) \iff \\ 1 + z + \bar{z} + z \bar{z} =1 + iz - i\bar{z} + z \bar{z} \iff \\ z + \bar{z} = i (z -\bar{z}) \iff \\ \frac{z + \bar{z}}{2} = \frac{i (z -\bar{z})}{2} = -\frac{z -\bar{z}}{2i} \iff \\ \mbox{re } z + \mbox{im z} = 0 $$

Switching to cartesian coordinates $z = x + i y$ with $x, y \in \mathbb{R}$ this is equivalent to $x + y = 0$ or as real valued linear function $$ y(x) = -x $$

Switching back to complex numbers, we can equivalently say that $z$ lies on the line $$ (1-i)t \quad (**) $$ with $t \in \mathbb{R}$.

Another way to interpret $(*)$ geometrically is \begin{align} |1 + z | &= |z - (-1 + 0 i)| = r_1 \\ |1 -i \bar{z}| &= |\overline{1 -i \bar{z}}| = |1 + i z|= |i||-i + z| = |z - (0+i)| = r_2 \end{align} The first equation describes a circle with center $(-1,0)$, the second a circle with center $(0,1)$. Both circles, if having the same radius $r$ each, intersect at two points, all intersection points form the line $(**)$ given above. Here is a plot for $r = 10$ and $r = 5$.

Answer 2

$|1+z|=|1-i\bar{z}|=|1+iz|=|-i+z|$. Hence $z$ has the same distance between $(-1,0)$ and $(0,i)$ . Thus $z$ has the form $t(-1+i), t\in \mathbb{R}$, lies on line $t(-1+i), t\in \mathbb{R}$.

Answer 3

John ZHANG provided a nice analytic description of the fact that this equation represents a line. Here's a command to visualize it with Mathematica, that works in Wolfram Alpha as well.

ContourPlot[Abs[(1+(x+I*y))/(1-I(x-I*y))]==1,{x,-3,3},{y,-3,3}]