All Questions
8
questions
7
votes
1
answer
177
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Product of differences of roots of unity
$\DeclareMathOperator{\lcm}{lcm}$
Denote by $\mu_n$ a primitive n-th root of unity. Examples suggest that
$$\prod_{0\le k\lt a \\ 0 \le l \lt b \\ k/a \neq l/b} (\mu_a^k - \mu_b^l)=\pm \lcm(a,b)^{\gcd(...
- 1,252
-3
votes
2
answers
101
views
Prove that a sum of squares to the $n$th power is also a sum of squares
Let $p$ and $q$ be real numbers, let $n$ be a positive integer, and let $i$ be the imaginary unit. Using the factorisation $p^2+q^2=(p+qi)(p-qi)$, prove for any $p$, $q$ and $n$, that the sum of the ...
1
vote
2
answers
1k
views
$z$ is a complex number such that $z^7=1$, where $z\not =1$. Find the value of $z^{100}+z^{-100} + z^{300}+z^{-300} + z^{500}+z^{-500}$
Let $z=e^{i\frac{2\pi}{7}}$
Then the expression, after simplification turns to
$$2[\cos \frac{200\pi}{7} +\cos \frac{600 \pi}{7} +\cos \frac{1000\pi}{7}]$$
How do I solve from here?
- 6,191
7
votes
1
answer
438
views
Which integer combinations of $n$-th roots of unity are zero?
Let $\omega$ be primitive $n$-th root of unity. Can we determine all tuples of integers $(c_1, c_2,\ldots,c_n) $ such that $$c_1+c_2 \omega + c_3 \omega^2+\cdots+ c_n \omega^{n-1}=0 \,?$$
It is clear ...
0
votes
2
answers
295
views
What are the factors of (2+i) in Z[i]?
The complex number $2+i$ factors as $i\cdot (1-2i)$ and $(-i)\cdot (1-2i)$. But those factorizations seem trivial. Are there any other ways to factor 2+i within the Z[i]?
7
votes
1
answer
212
views
A curious property of exponential sums for rational polynomials?
An article led me to generate some graphs of exponential sums of the form $S(N)=\sum_{n=0}^Ne^{2\pi i f(n)}$, where $f(n)= {n\over a}+{n^2\over b}+{n^3\over c}$ with $a,b,c\in\mathbb{N}_{>0},\,$ ...
- 15.1k
5
votes
2
answers
213
views
Is it true that $(a^2-ab+b^2)(c^2-cd+d^2)=h^2-hk+k^2$ for some coprime $h$ and $k$?
Let us consider two numbers of the form $a^2 - ab + b^2$ and $c^2 - cd + d^2$ which are not both divisible by $3$ and such that $(a, b) = 1$ and $(c,d) = 1$. Running some computations it seems that ...
- 51
3
votes
0
answers
210
views
Apply Cardano's method to find the roots of $f(x)=x^3 - 6x + 4$
Apply Cardano's method to find the roots of $f(x) = x^3 − 6x + 4$
There was a part a) to this question with $f(x)=x^3 - 6x - 6$ which yielded a positive number under the square root using Cardano's ...
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