I have tried this complex analysis problem for hours on my own but am still stumped.     Suppose $\zeta_7 = e^{2\pi i/7} = \cos(2\pi/7)+i\sin(2\pi/7)$. Prove that $$\sum_{k=0}^{6} \zeta_7^k = 0.$$   I have so far obtained that $$\sum_{k=0}^{6} (\zeta_7)^k = 1 + (\zeta_7) + (\zeta_7)^2 + (\zeta_7)^3 +(\zeta_7)^4 +(\zeta_7)^5 + (\zeta_7)^6,$$ but that's as far as I get.
21995
Question: If $c\neq 1$ is an $n^{th}$ root of unity then, $1+c+...+c^{n-1} = 0$  Attempt: So I have established that I need to show that $$\sum^{n-1}_{k=0} e^{\frac{i2k\pi}{n}}=\frac{1-e^{\frac{ik2\pi}{n}}}{1-e^{\frac{i2\pi}{n}}} =0$$ by use of the sum of geometric series'. My issue is proceeding further to show that this is indeed true.
47861
I know some nice ways to prove that $\zeta(2) = \sum_{n=1}^{\infty} \frac{1}{n^2} = \pi^2/6$.  For example, see  or the answers to the question "?"  Are there any nice ways to prove that $$\zeta(4) = \sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{90}?$$  I already know some proofs that give all values of $\zeta(n)$ for positive even integers $n$ (like #7 on Robin Chapman's list or Qiaochu Yuan's answer in the linked question).  I'm not so much interested in those kinds of proofs as I am those that are specifically for $\zeta(4)$.  I would be particularly interested in a proof that isn't an adaption of one that $\zeta(2) = \pi^2/6$.
104317
Let $z_1,\dots,z_n$ be non-zero complex numbers such that $\sum_{k=1}^n \frac{1}{z_k} = 0$. Prove that for any line $ax+by=0$ passing through the origin, $z_1,\dots,z_n$ cannot all lie in either of the half-spaces $\{ ax+by&lt;0\}$, $\{ ax+by &gt;0\}$. Any help would be greatly appreciated, thanks in advance!
35974
I found a solution but when testing it, it does not work....?    $$\cos z + \sin z = i \\ \frac{e^{iz} + e^{-iz}}{2} + \frac{e^{iz} - e^{-iz}}{2i} = i \\ e^{2iz}(1+i) + 2e^{iz} + i - 1 = 0 \\ e^{iz} = \frac{-1 \pm \sqrt{2-i}}{1+i} \\ z = -i \mathrm{Log}\left(\frac{-1\pm\sqrt{2-i}}{1+i}\right) + k2\pi$$    Upon testing it in W|A with $k=0$, I don't get the required result $(=i)$. I'm not sure where I went wrong and I did this a few times already.
19908
For what complex values of $z$ does the following sum converge?  $$\sum_{k=0}^\infty {\frac{z(z-1)\cdots(z-k+1)}{k!}} $$  And how would you prove it?  Mathematica seems to suggest the sum converges as long as $\Re(z) \ge 0$, regardless of the imaginary part.  Is that right?
97412
I'm looking for a way to prove $$\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^{2m+1}}=\frac{(-1)^m E_{2m}\pi^{2m+1}}{4^{m+1}(2m)!}$$ I know that $$\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^{2m+1}}=\frac{1}{4^{2m+1}}\left(\zeta\left(2m+1,\frac14\right)-\zeta\left(2m+1,\frac34\right)\right)$$ so maybe I could simplify the above more?
54597
How to find the real part of the complex number (in Euler's form) $ z = e^{e^{i \theta } } $ ?  I got confused on how to proceed.  I am a beginner to complex numbers.
32642
Can somebody help me to prove this identity?     $$\prod_{k=1}^{n-1} \left(\;\cos\theta-\cos\frac{k\pi}{n}\;\right) = 2^{1-n}\sin(n\theta)\csc\theta$$   I was thinking of using the properties of the complex numbers and the binomial theorem, because I found this problem in the Complex Variables book from Schaum's, but it did not solve my problem.  Does anyone have an idea?  Thanks in advance!
19167
$$\zeta(it)=2it\pi it−1\sin(i\pi t/2)\Gamma(1−it)\zeta(1−it).$$ Everything on the RHS is never zero,  Does that means LHS has no zeros, since $\sin(s)$ has a simple zero at $s=0$ while $\zeta(1−s)$ has a simple pole at $s=0$  (the Laurent expansion), so the product $\sin(i\pi t/2)\zeta(1−it)$ is finite and nonzero at $t=0$.  My question is that from functional equation one can't get direct answer $\zeta(0)=−1/2$ unless using alternate infinite series !
81871
I recall from complex analysis a while ago that the series $$\sum_{n=1}^\infty \frac{z^{n!}}{n}$$ has radius of convergence 1, diverges on roots of unity, and converges elsewhere on the circle. It's simple enough to see that it has to diverge on roots of unity: consider $z$ such that $z^k = 1$ for some $k \in \mathbb{N}.$ Plug this into the series and you see that past the $k$'th term, $z^{n!} = (z^k)^m = 1$ for some $m \in \mathbb{N}$; hence, the series becomes harmonic and must diverge.  But I don't recall how to show that the series converges for other values on the circle. If this is simple I'd like just a hint, if it is involved I would prefer a sparse outline of the proof. Thank you!
9512
Can anyone please explain to me how to prove that Riemann zeta function is 0 for all negative even numbers. In many references , they have just given the statement without any proof. Any explanation or hint will be a great help. Thanks in advance.
54876
Let $\zeta_7= e^{i2\pi/7}$ be the $7$th root of unity.Find the quadratic sub field of $\mathbb Q(\zeta_7)$ which can be expressed in the form $\mathbb Q(\sqrt D)$,where $D$ is an integer.  My try:let $\omega=\zeta_7+\zeta_7^{-1}\in\mathbb{Q}(\zeta_7)$ is real which is $\omega=2\cos(2\pi/7)$. Thus $K=\Bbb{Q}(\omega)$ is a proper subfield of $\Bbb{Q}(\zeta_7)$. And the polynomial $$ p(x)=(x-\zeta_7)(x-\zeta_7^{-1})=x^2-\omega x+1 $$ has its coefficients in the subfield $K$. Thus it has to be the minimal polynomial of $\zeta_7$, and $[K(\zeta_7):K]=2$.  But the problem is that what is $D?$ and how can i find that?Thank you
107381
We know that $1+\alpha_1+\alpha_2+...+\alpha_{n-1}=0$ where $\alpha_i$ are the roots of $z^n=1$. How can I prove that: $$1+\sum_{i=1}^{n-1}\alpha_i^m=\begin{cases}0\quad m\in Z,m\not\equiv0\pmod n\\n\quad m\in Z,m\equiv0\pmod n\\\end{cases}$$
46297
Find $$M:=\sum_{n=1}^{\infty}\tan^{-1}\frac{2}{n^2}$$   There's a solution  that uses complex numbers which I didn't understand and I was wondering if the following is also a correct method.   My proposed solution   $$\begin{align} &amp;\sum_{n=1}^{\infty}\tan^{-1}\frac{2}{n^2}\\ =&amp;\sum_{n=1}^{\infty}\tan^{-1}\frac{(1+n)+(1-n)}{1-(1+n)(1-n)}\\ =&amp;\sum_{n=1}^{\infty}(\tan^{-1}(1+n)+\tan^{-1}(1-n))\\ =&amp;\sum_{n=1}^{\infty}(\tan^{-1}(n+1)-\tan^{-1}(n-1)) \end{align} $$  And this implies  $$M=\lim_{m\to\infty}(\tan^{-1}(m+1)+\tan^{-1}m-\tan^{-1}1-\tan^{-1}0)=\frac{3\pi}{4}$$
40295
I'd like to see a proof of this identity I discovered:  $$ \sum_{i=1}^{n}\sum_{j=1}^{i}a_j = \sum_{i=1}^{n}a_i(n-i+1) $$  Probably there's something I forgot because I couldn't manage it.  [edit]Possibly with just identities, not examples please!  [edit2]No induction procedure, either.
79764
I would to prove that  $$1+\cos \theta+\cos 2\theta+\ldots+\cos n\theta =\displaystyle\frac{1}{2}+  \frac{\sin\left[(2n+1)\frac{\theta}{2}\right]}{2\sin\left(\frac{\theta}{2}\right)}$$ given that  $$1+z+z^2+z^3+\ldots+z^n=\displaystyle\frac {1-z^{n+1}}{1-z}$$ where $z\neq 1$.  I put $z=e^{i\theta}$. I already got in left hand side cos exp in real part, but there is a problem in the right hand side, I can't split imaginary part and real part. Please help me. Thanks in advance.
111145
Related to the question :  It is well known that   $$\zeta(2n) = f(2n) \pi^{2n}$$  where $f(n)$ is an function in rationals: (the denominator = OEIS : 6, 90, 945, 9450).  Apery proved that $\zeta(3)$ is irrational, and presumably $f(2n+1)$ is so for all (but the latter is open).  But what else is known about $f(2n+1)$? What sort of pattern does it share (if any) with $f(2n)$? Numerically does it fit well? Is there any clue/progress since Apery as to its form? Of course, $f(2n+1)$ may in effect divide out the $\pi$, meaning that somehow $\pi$ is not involved in $\zeta(2n+1)$ in any essential way.