All Questions
28
questions
56
votes
7
answers
8k
views
Can we calculate $ i\sqrt { i\sqrt { i\sqrt { \cdots } } }$?
It might be obvious that $2\sqrt { 2\sqrt { 2\sqrt { 2\sqrt { 2\sqrt { 2\sqrt { \cdots } } } } } } $ equals $4.$ So what about $i\sqrt { i\sqrt { i\sqrt { i\sqrt { i\sqrt { i\sqrt { \cdots } } } } } } ...
user536517
9
votes
2
answers
356
views
computing $A_2=\sum_{k=1}^{n}\frac{1}{(z_k-1)^2} $ and $\sum_{k=1}^n \cot^2\left( \frac{k\pi}{n+1}\right)$
Assume that $z_1,z_2,...,z_n$ are roots of the equation $z^n+z^{n-1}+...+z+1=0$.
I was asked to compute the expressions
$$A_1=\sum_{k=1}^{n}\frac{1}{(z_k-1)} ~~~~~~and~~~~~~A_2=\sum_{k=1}^{n}\...
- 24.2k
0
votes
1
answer
74
views
Trigonometry and Complex Numbers with Series [duplicate]
The number
$$\text{cis}75^\circ + \text{cis}83^\circ + \text{cis}91^\circ + \dots + \text{cis}147^\circ$$ is expressed in the form $r \, \text{cis } \theta,$ where $0 \le \theta < 360^\circ$. ...
- 61
6
votes
2
answers
2k
views
Prove that $\cos \frac{2\pi}{2n+1}+\cos \frac{4\pi}{2n+1}+\cos \frac{6\pi}{2n+1}+...+\cos \frac{2n\pi}{2n+1}=\frac{-1}{2}$
Prove that $$\cos \frac{2\pi}{2n+1}+\cos \frac{4\pi}{2n+1}+\cos \frac{6\pi}{2n+1}+...+\cos \frac{2n\pi}{2n+1}=\frac{-1}{2}$$
My attempt,
Let an equation $x^{2n+1}-1=0$, which has roots $$\cos \frac{...
- 7,798
10
votes
3
answers
354
views
Finding $\sum_{k=0}^{n-1}\frac{\alpha_k}{2-\alpha_k}$, where $\alpha_k$ are the $n$-th roots of unity
The question asks to compute:
$$\sum_{k=0}^{n-1}\dfrac{\alpha_k}{2-\alpha_k}$$
where $\alpha_0, \alpha_1, \ldots, \alpha_{n-1}$ are the $n$-th roots of unity.
I started off by simplifiyng and got ...
- 7,551
0
votes
3
answers
2k
views
If $\omega = e^{(\frac{2\pi i}{n})}$ why $1+ \omega + \omega^{2} + ... + \omega^{n-1} = 0 $? [duplicate]
Let $\omega = e^{(\frac{2\pi i}{n})}$ why $1+ \omega + \omega^{2} + ... + \omega^{n-1} = 0 $?
I saw this on a algebra PPT slice. However the teacher did not explain why this equation is correct, can ...
- 269
7
votes
2
answers
305
views
Bounding a sum involving a $\Re((z\zeta)^N)$ term
This is a follow up to this question. Any help would be very much appreciated.
Let $k\in\mathbb{N}$ be odd and $N\in\mathbb{N}$. You may assume that $N>k^2/4$ or some other $N>ak^2$.
Let $\...
- 7,789
2
votes
2
answers
185
views
Simplifying this (perhaps) real expression containing roots of unity
Let $k\in\mathbb{N}$ be odd and $N\in\mathbb{N}$. You may assume that $N>k^2/4$ although I don't think that is relevant.
Let $\zeta:=\exp(2\pi i/k)$ and $\alpha_v:=\zeta^v+\zeta^{-v}+\zeta^{-1}$.
...
- 7,789
3
votes
1
answer
64
views
What is condition that the sum of $n$ complex numbers eaquals their product
Let $n\geq2$ and let $\{z_1,\dots,z_n\}$ be a set of complex numbers.
Is there a condition on the $z_i$'s such that
$$\sum_{i=1}^n z_i=\prod_{i=1}^n z_i$$
is identically true?
For $n=2$ the ...
- 1,008
3
votes
2
answers
175
views
Evaluate $S=\left|\sum_{n=1}^{\infty} \frac{\sin n}{i^n \cdot n}\right|$
Evaluate
$$ S=\left|\sum_{n=1}^{\infty} \dfrac{\sin n}{i^n \cdot n}\right|$$
where $i=\sqrt{-1}$
For this question, I did the following,
Let
$$
\begin{align*}
S &= \sum_{n=1}^{\infty} \...
user208998
1
vote
5
answers
1k
views
Sequence with imaginary numbers
How do I solve this:
$$1 - i + i^2 - i^3 + i^4 + ... + i^{100} - i^{101}$$
I see that any 4 consecutive members of the sequence equal $0$. If I extract $1$ and $-i^{101}$, I see there are 100 ...
- 13
3
votes
3
answers
129
views
Series Summation: $\sum\limits_{k=1}^{N-1}\frac{1}{z-w_k}$ where $w_k=e^{\frac{2\pi i k}{N}}$
I have the series
$$\sum_{k=1}^{N-1}\frac{1}{1-w_k} $$ where $w_k=e^{\frac{2\pi i k}{N}}$, how can I find the summation of this , another question related to this sum
$$\sum_{k=1}^{N-1}\frac{1}{z-w_k} ...
- 41
2
votes
1
answer
399
views
A finite sum of trigonometric functions
By taking real and imaginary parts in a suitable exponential equation, prove that
$$\begin{align*}
\frac1n\sum_{j=0}^{n-1}\cos\left(\frac{2\pi jk}{n}\right)&=\begin{cases}
1&\text{if } k \...
- 33