The set of sums of unit complex numbers

Question

What are the properties of the set of complex numbers that can be expressed as a finite sum of distinct unit complex numbers? (Unit complex numbers are complex numbers with absolute value 1.)

(Sorry my english is not perfect)

It is trivial that every complex number is expressible as a sum of finitely many unit complex numbers, but as far as I am aware, the sum contains some elements more than once.

Can all complex numbers also be constructed from a finite sum of points on the unit circle, each point only counting once? If not, what does the set of these sums look like? What subsets does it contain? What are some numbers not in the set?

Answer 1

Yes, such construction is possible.

Hint: consider real numbers first, and show that any $x \in \mathbb{R}$ can be written as a finite sum of distinct real terms $x=x_1+x_2+\ldots+x_n$ where $|x_k| \lt 2\,$. Then express each term as $x_k = z_k + \bar z_k$ where $z_k$ is a complex number on the unit circle, in fact a root of the equation $z^2 - x_k z + 1 = 0$ which has complex roots since $\Delta = x_k^2-4 \lt 0$, and the roots $z_k, \bar z_k$ lie on the unit circle since $|z_k|^2=z_k\bar z_k=1$. It follows that any real $x$ can be written as a sum of distinct complex numbers on the unit circle. Next, repeat for $y$ with some care to not reuse the same numbers on the unit circle, then add $x+iy$ for an arbitray complex $z=x+iy$.

Answer 2

Let $r$ be a positive real number and let $k$ be the smallest integer strictly greater than $\max(r,1)$. Visualise a chain of $k$ unit intervals hanging freely (gravity acts in the direction of the negative imaginary axis), with one end at the origin and the other at $r$. It is clear that all of these intervals, if interpreted as complex numbers, will have modulus $1$, they will all be pointing in different directions so they all have different arguments, and their sum is $r$.

To achieve the same thing for $z=re^{i\theta}$, multiply all these intervals by $e^{i\theta}$. They still have modulus $1$ and different arguments, and they add up to $z$.