Any hints?
If z = \cos \frac{2 \pi}{n} + i \sin \frac{2 \pi}{n}, n \in \mathBB{N}; n \ge 3 define
A = \{1,z,z^2,z^3,...,z^{n-1} \}
B = \{ 1, 1+z, 1+z+z^2,...,1+z+z^2+...+z^{n-1} \}
Determine A \bigcap{B}
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8 Answers
66The elements of A are obviously the n-th roots of unity.
The element 1 is common to both. We shall prove that it is the only one.
For each k =2, 3, . . .,n, we have
1+z+z^2+\ldots +z^{k-1}=\dfrac{1-z^k}{1-z}
Hence
|1+z+z^2+\ldots +z^{k-1}|=\dfrac{|1-z^k|}{|1-z|}
=\dfrac{2(1-\cos(2k\pi/n))}{2(1-\cos(2\pi/n))}=\dfrac{\sin^2(k\pi/n)}{\sin^2(\pi/n)}\neq 1
So none of the elements in set B (except 1) has modulus 1 but each element of set A has a modulus 1.
So A\cap B =\{1\}
66Geometrically, the sum of two vectors has the same length as the original ones only if the angle between them is 120°. This happens only for n = 3 in the above case. But in that case, the elements in A are 1, ω, ω2, while that in B are 1, -ω2, 0. So the only common element is 1.
For n > 3, none of the elements in set B (except 1) lie on the unit circle.
341Thats not correct. check your result for the 4th roots of unity
66Yes, that's true. I have committed a mistake. Even in the post #5, I need to take a square root.
66Okay, we require those values of k in {1,2,3,...n} such that
|1+z+z^2+\ldots+z^{k-1}|=1
which gives
\cos\left(\dfrac{2k\pi}{n}\right)=\cos\left(\dfrac{2\pi}{n}\right)
whose general solution is
\dfrac{2k\pi}{n}=2m\pi \pm \dfrac{2\pi}{n}
which gives
\dfrac{k\mp 1}{n}=m
that is k ± 1n should be an integer. This gives two values of k, namely, 1 and n-1.
But for k =n-1, we have 1+z+z^2+\ldots +z^{n-2}=-z^{n-1}
So if -z^{n-1} is to coincide with one of the n-th roots, n must be even. To see this, we see that if n = 2m, then
-z^{n-1}=e^{i\left(\pi+\frac{2(2m-1)\pi}{2m}\right)}=e^{i\frac{6m-2}{2m}\pi}=e^{i\left(2\pi + \frac{2m-2}{2m}\pi\right)}=e^{i\frac{2m-2}{2m}\pi}=e^{i\frac{2(n/2-1)}{n}\pi}
which belongs to A. So,
A\cap B =\begin{cases} \{1\} & \text{ if } n=\text{ odd} \\[2ex] \{1,\ 1+z+z^2+\ldots + z^{n-2}\} & \text{ if } n=\text{ even} \end{cases}