1

I found this problem in a text-book, no solution offered. I'm curious because it seems like a very interesting result. Full statement is:

Let $M \subseteq \mathbb{C}$, a set with the following properties:
1. if $x\in{\mathbb{C}}$ with $|x|=1$, then $x \in{M}$
2. if $x=a_1+a_2$ and $a_1,a_2 \in{M}$, then $x \in{M}$

Show that $M=\mathbb{C}$.

Any suggestions are welcome, thanks in advance :)

bof
  • 6,156
Lisa
  • 733
  • 3
    A hint on one possible approach: $e^{i\theta}$ and $e^{-i\theta}$ both have modulus 1, and their sum is $2\cos\theta$, which means that any real number between -1 and +1 can be obtained as a sum of two such numbers. – Old John Oct 20 '13 at 08:49

3 Answers3

2

An analytic proof using Old John's hint.

Let $z=r\,e^{i\phi}$ with $r\ge0$ and $0\le\phi<2\,\pi$. Let $[r]$ be the integer part of $r$ and $\{r\}$ its fractional part. Then $$ z=e^{i\phi}+\dots([r]\text{ times})\dots+e^{i\phi}+\{r\}\,e^{i\phi}. $$ It is enough to show that the las term in the sum is the sum of complex numbers of modulus $1$. Let $\theta=\arccos(\{r\}/2)$. Then $$ \{r\}\,e^{i\phi}=(e^{i\theta}+e^{-i\theta})\,e^{i\phi}=e^{i(\phi+\theta)}+e^{i(\phi-\theta)}. $$

Julián Aguirre
  • 76,354
1

It is geometrically equal to that between any two points on a plane, one can draw a chain of unit-radius circles, the centre of each is on the circumference of the previous. Eventually they must intersect. The plane becomes the argand diagram.

Suppose one starts with a point $a+bi$. It is possible to draw a circle to find a new point $a-1 + bi$, and steadily reduce the real and imaginary components to less than unity. One then has a point $A+Bi$, lying in the unit circle. A circle drawn around this point, will then intersect the unit circle at some $c+di$. A circle drawn at $c+di$ will pass through $A+Bi$, and the chain can be followed back to $a+bi$.

wendy.krieger
  • 6,856
1

Think geometrically. Begin at any point in the complex plane, and draw a unit radius circle centered at your point. Pick a point on the circle closer to the origin, and use it as the center of another unit circle. Repeat this process until you have a chain of unit circles, the last one intersecting the unit circle centered at the origin.

Now, connect the centers of these circles from the origin out to your point. Voilà. Your complex number exhibited as a sum of unit modulus complex numbers.

Sammy Black
  • 25,273
  • while it is an intuitive way to express an equivalent problem, it doesn't seem like a very rigorous proof – Lisa Oct 20 '13 at 08:57
  • 1
    @Andreea: Much communication is implicit: I would consider this very rigorous if it was given to me by one of my colleagues. But from a student who is being tested on his ability to formulate inductive arguments from proof sketches, I would demand more detail. –  Oct 20 '13 at 10:00