Bounded set congruent to its subset It is a great problem to find a bounded set which is congruent to its subset. My friend told me that the set $z^n$ works where $z$ is a complex number, for which $|z|=1$ and $z$ is not a root of unity, and where $n\in \mathbb{N}$.
However, I only started learning about complex numbers, can anyone explain why this works? Or is there a better example? Thanks in advance!
 A: Your friend is probably referring to a fact about the geometric interpretation of complex numbers. Addition and multiplication of numbers has some geometric significance which allow easy expression of congruence of shapes if we represent them within the complex plane.
If you imagine the complex plane in the usual way, with perpendicular axes for the real and imaginary parts of the number and with $i$ being $90^{\circ}$ counterclockwise from $1$, you can first observe that a function
$$f(z)=z+c$$
represents a translation where each point is shifted by $c$, which is regarded as a vector from the origin. If you applied this function to each point of a shape, its image would be a congruent shape somewhere else in the plane.
Similarly, multiplication represents a dilative rotation about the origin - that is, it scales and rotates by some amount. For instance, if $s$ is real, then
$$f(z)=sz$$
scales the whole plane by a factor of $s$, sending shapes to similar ones. The function
$$f(z)=iz$$
takes the whole plane and rotates it by $90^{\circ}$ counterclockwise - so takes shapes to congruent ones. In general, if $\alpha$ is a complex number the function
$$f(z)=\alpha z$$
takes points and rotates them about the origin by the argument of $\alpha$ (the angle the vector $\alpha$ makes with the positive real line) and scales by the magnitude of $\alpha$ (how far it is from the origin). So, $f(z)=\alpha z$ describes a rotation whenever $|\alpha| =1$ (i.e. when $\alpha$ lies on the unit circle).
This in hand, we can find out that any function of the form
$$f(z)=\alpha z+c$$
where $|\alpha|=1$ is either a pure rotation about some point or a translation (if $\alpha = 1$) - which are both isometries of the plane, meaning that they take congruent shapes to congruent shapes. In fact, every orientation preserving isometry of the plane is of this form.
To get the whole set of isometries of the plane, one also considers reflections, which can be obtained from the complex conjugate - which is reflection over the real line, defined by $\overline{a+bi}=a-bi$. Functions of the form
$$f(z)=\alpha \bar z + b$$
where $|\alpha|=1$ are the orientation reversing isometries, corresponding to reflections and glide reflections of the plane.
So, to say that two sets $S$ and $T$ are congruent means that there is some function $f$ of one of the above two forms such that the image of $S$ under $f$ is $T$ (that is, if $f(S)=T$).
Your friend is stating that if $|\alpha|=1$ then the set $\{1,\alpha,\alpha^2,\alpha^3\}$ is bounded (since every element has norm $1$) and is congruent to a proper subset of itself (in particular the subset $\{\alpha,\alpha^2,\alpha^3,\ldots\}$). You can see this because if you consider the function $f(z)=\alpha z$, which is some rotation of the plane about the origin, we have $f(\{1,\alpha,\alpha^2,\ldots\}) = \{\alpha,\alpha^2,\alpha^3,\ldots\}$.
This is the same as if you, just in the real plane, marked out a circle $C$ and then chose some angle $\theta$ and marked a point, then a point $\theta$ counterclockwise of that, then $2\theta$ counterclockwise of that and so on. Rotating that set of points counterclockwise by $\theta$ would yield a subset of the original set - and this subset is proper so long as no multiple of $\theta$ ever comes back to the starting point, which occurs if $\frac{\theta}{2\pi}$ is irrational. Similarly, the set $\{\alpha,\alpha^2,\ldots\}$ is a proper subset of $\{1,\alpha,\ldots\}$ whenever $\alpha^n$ is never $1$ for any $n>0$ - meaning that $\alpha$ is not a root of unity.
It's worth noting that this trick works with any isometry $f$: If you let $f$ be an isometry and $p$ be any point, then $\{p,f(p),f(f(p)),\ldots\}$ is congruent to $\{f(p),f(f(p)),f(f(f(p)))\}$ by $f$. It just happens that whenever $f$ is a pure rotation by an angle that is not a rational multiple of $2\pi$ and $p$ is a point that is not the center of rotation, this yields a bounded set that is congruent to a proper subset of itself - and, in fact, a little more work shows that if a bounded set $S$ is congruent to a proper subset $T$ of itself by a symmetry $f$, then $f$ had to have been such a pure rotation.
