I have $\frac{z^6}{(z+1)^6} = 1 = e^{2k\pi i}$.
15348
Find all $x$ such that $$x^6=(x+1)^6.$$   So far, I have found the real solution $x= -\frac{1}{2}$, and the complex solution $x = -\sqrt[3]{-1}$.  Are there more, and if so what/how would be the most efficient find all the solutions to this problem? I am struggling to find the rest of the solutions.
64569
I have to show that if $|z| &lt; 1$, $z \in \mathbb{C}$, $$(1+z)\prod_{n=1}^{+\infty}(1+z^{2^{n}})= \frac{1}{1-z}$$  I want to understand how to do this kind of exercises, any hint ?
30689
I have one question about a proof in Rudin's book:  If $z_1 ..., z_N$ are complex numbers then there is a subset $S$ of  $ \{1,..., N \}$ for which $|\sum_{k \in S} z_k| \ge \frac{1}{\pi} \sum_{1}^N |z_k|$  Proof:  Write $z_k = |z_k| e^{i \alpha_k}$.   For $\phi \in [- \pi, \pi] $ let  $\ \ S(\phi):= \{ k | \cos (\alpha_k - \phi)&gt;0 \}$.   Then $|\sum _{S(\phi)} z_k| = |\sum_{S(\phi) } e^{-i \phi} z_k| \ge Re(\sum_{S(\phi) } e^{-i \phi} z_k) = \sum_1^N |z_k| \cos ^+ (\alpha_k - \phi) $.  Could you tell me why the two equations hold (the inequality is quite obvious) and what $\cos ^+$ stands for?
38052
I understand why $e^{i\pi} = -1$and as a result  $e^{i2\pi}=\left(e^{i\pi}\right)^2=1.$ These results can be confirmed using Euler's formula  But why does $e^{i\pi/3}\neq -1$ as we can write it $(e^{i\pi})^{1/3}$ and as such $(-1)^{1/3}$ the cubic root of $-1$ is $-1$ itself.  Yet using Euler's formula we get $e^{i\pi/3}=\frac{1}{2}+i\frac{\sqrt 3}{2}$.
112775
Prove that   $$\sum_{n=1}^\infty\frac1{n^6}=\frac{\pi^6}{945}$$ by the Fourier series of $x^2$.   By Parseval's identity, I can only show $\sum_{n=1}^\infty\frac1{n^4}=\frac{\pi^4}{90}$. Could you please give me some hints?
59129
If $|z|=1$, show that: $$\mathrm{Re}\left(\frac{1 - z}{1 + z}\right) = 0$$   I reasoned that for $z = x + iy$, $\sqrt{x^2 + y^2} = 1\implies x^2 + y ^2 = 1$ and figured the real part would be:  $$\frac{1 - x}{1 + x}$$  I tried a number of manipulations of the equation but couldn't seem to arrive at any point where I could link the two to show that the real part was = 0.
29853
As the title suggests, I would like to show that: $$\frac{1}{2\pi}\int^{2\pi}_{0}(2\cos\theta)^{2n}d\theta=\frac{(2n)!}{n!n!}$$ for every positive integer $n$.     How can I go about doing that? I understand the that there are already solutions for the problem: $\frac{1}{2\pi}\int^{2\pi}_{0}(\cos\theta)^{2n}d\theta= {2n \choose n} \frac{\pi}{2^{2n-1}}$. Is there any modifications I can do to obtain the desired solution?
114177
We know that $$\int_0^1\frac{\ln (1+x)}{1+x^2}dx=\frac{\pi}8\ln 2,$$ but how about $$\int_0^1\frac{x\ln (1+x)}{1+x^2}dx?$$Prove that $$\int_0^1\frac{x\ln (1+x)}{1+x^2}dx=\frac{\pi^2}{96}+\frac{\ln^2 2}{8}.$$
90354
$$\frac{1}{z-\sin z}$$ $$\frac{1}{z(1-\frac{\sin z}{z})}$$  the laurent's transform is also used to simplify the function.  the residue is found a $\frac{2*36*2}{5}$ but the actual answer is given as $\frac{3}{10}$ the residue is given as The solution is incorrect and also the pole is of the order 3.
105702
I know that $$\int^\infty_0\frac x{e^x-1}dx=\frac{\pi^2}{6}$$ For substituting $u=2$ into $$\zeta(u)\Gamma(u)=\int^\infty_0\frac{x^{u-1}}{e^x-1}dx$$  However, I suspect that there is an easier proof, maybe by the use of complex analysis. I haven't learnt the  yet. All I know is the above formula and $\zeta(2)=\frac{\pi^2}{6}$. Can we can use the above integral to find out some of the $\zeta$'s value?
34341
What's wrong with this : $$e^{i\pi} = -1$$ $$\therefore e^{2i\pi} = 1$$ $$\therefore \log \left( e^{2i\pi} \right ) = \log(1) = 0$$ $$\therefore 2i\pi = 0$$
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?
10671
I've found this duplicate from '15: , but I think it's wrong, since it looks for the Laurent expansion in ${|z-1|&gt;1}$.   After developing it by myself, I've reached the following:  $$f(z)=\frac{e^z}{z-1}=\frac{1}{z-1}\cdot e^z$$  Then, $e^z=\sum_{n\geq 0} \frac{z^n}{n!}$ and $$\frac{1}{z-1}=\sum_{n \geq 0} \frac{1}{z^{n+1}}$$.  Here I thought of using the Cauchy Product, but I end up with  $$f(z)=\sum_{n \geq 0} \sum_{k=0}^n \frac{z^{2k-n-1}}{k!}$$ (How can I make this font bigger? The exponent is too small. Sorry about that. The exponent would be $2k-n-1$).  So I have to sum whenever $2k-n-1 = 1$  If I'm not getting it wrong, this means that I have to calculate $$\sum_{n \geq 0} \frac{1}{(1+2n)!}$$ (I'm not sure how to show it better, but I've put the first 3 terms and were $\frac{1}{1!}, \frac{1}{3!}$ and $\frac{1}{5!}$. I'm not sure how to calculate that. The nearest thing I thought was that $$\sum_{n \geq 0} \frac{1}{n!}=e $$. Maybe what I found before could be considered as an answer, but I think I can go one step further.  Thanks
88667
Let $z_1 = x_1 +iy_1$  and $z_2 = x_2 +iy_2$   be two complex numbers. The dot product of $z_1$ and $z_2$ is defined by $\langle z_1 , z_2 \rangle = x_1x_2+y_1y_2$ For non zero $z_1$ and $z_2$ prove the following  $ \langle z_1, z_2 \rangle = |z_1||z_2|\cos\theta$ (where $0 \le \theta \le \pi$ is the angel between $z_1$ and $z_2$.) $\langle z_1, z_2 \rangle$ = $Re(\overline{z_1}z_2)$ $\langle z_1, z_2 \rangle$ = $\frac{1}{2}(\overline{z_1}z_2 + z_1\overline{z_2})$ $z_1 ⊥ z_2 \iff \theta = \frac{\pi}{2}$
102770
Show that  $ \quad \quad z^5-1=(z-1)(z^2+2z\cos{\dfrac{\pi}{5}}+1)(z^2-2z\cos{\dfrac{2\pi}{5}}+1)$   and deduce from this the closed formulas for $\cos{\dfrac{\pi}{5}}$ and $\cos{\dfrac{2\pi}{5}}$.    Here's what I did:  I assumed that it was correct, and using the fact that $z^5-1=(z-1)(z^4+z^3+z^2+z+1)$, I said that:  $ \quad z^4+z^3+z^2+z+1=(z^2+2z\cos{\dfrac{\pi}{5}}+1)(z^2-2z\cos{\dfrac{2\pi}{5}}+1)$  Letting $a=\cos{\dfrac{\pi}{5}}$ and $b=\cos{\dfrac{2\pi}{5}}$  $ \quad z^4+z^3+z^2+z+1=z^4+(2a-2b)z^3+(2-4ab)z^2+(2a-2b)z+1$   Equating the coefficients:  $\quad a-b= {1 \over 2} \quad$ and $ \quad ab={1 \over 4}$  Simultaneously solving, and knowing that $0&lt;a&lt;1$ and $0&lt;b&lt;1$:  $\cos{\dfrac{\pi}{5}}=\dfrac{1}{4} + \dfrac{\sqrt{5}}{4} \quad$ and $\quad \cos{\dfrac{2\pi}{5}}=-\dfrac{1}{4} + \dfrac{\sqrt{5}}{4}$  Then I said that, if I can prove the cosine formulas above in a different way, I have indirectly shown that the relationship shown in the question was true.  So, since $\cos{\pi}= \cos{(5 × (\frac{\pi}{5}))} =-1$ and with a little trig indentity,  $\quad 16{\cos^5{\pi \over 5}}-20{\cos^3{\pi \over 5}}+5{\cos{\pi \over 5}}+1=0$  With substitution $x=\cos{\pi \over 5}$:  $16x^5-20x^3+5x+1=0$, a polynomial that has roots at $x=-1$, $x= \frac{1}{4} - \frac{\sqrt{5}}{4}$, and $x=\frac{1}{4}+\frac{\sqrt{5}}{4}$.  The only possible solution is $\cos{\pi \over 5}=\frac{1}{4}+\frac{\sqrt{5}}{4}$, and the double angle formula shows that $ \cos{\frac{2\pi}{5}}=-\frac{1}{4} + \frac{\sqrt{5}}{4}$  My question, is there a simpler or more straightforward way of proving this?
56694
we are given $z^n=(z+1)^n=1$, $z$ complex number. we want to prove that $n$ is divisible by $6$. I showed that $|z|=|z+1|=1$. Hence $z$ is on the intersection of two unit circles, one centered at $(0,0)$ and the other one centered at $(-1,0)$. Then $z$ can take two values $z=\operatorname{cis}(2 \pi /3)$ or $z=\operatorname{cis}(4 \pi/3)$. Then from $z^n=1=\operatorname{cis}(2 \pi k)$, $k$ integer I showed that $n$ should be divisible by $3$. I am left to show that $n$ should be divisible by $2$. I think if I figure out a way to show that $\operatorname{arg}(z+1)=\pi/3$ or $-\pi/3$ then I would be done, by using $(z+1)^n=\operatorname{cis}(2\pi k)$. but how can I do this? Any ideas are welcome! Thanks in advance!
40466
If z is a complex number and $ \frac {z^2 + z+ 1} {z^2 -z +1}$   is purely  real then find the value of $|z|$ .  I tried to put $ \frac {z^2 + z+ 1} {z^2 -z +1} =k $ then solve for $z$ and tried to find |z|, but it gets messy and I am stuck. The answer given is |z|=1