Essentially, I need to show that e^inx is dense in the complex unit circle for irrational angle x. I'm not sure how to go about this but I think I need to show that between any two points on the unit circle, we can find some e^inx.
123376
I think the title of the question says it all. I unfortunately did not seem to conclude anything. Some ideas I had:  It is easy to show that (given $T$ is the rotation) $\{T^n(\theta)\}$ is a set of distinct points. Furthermore, given that the circle is a compact metric, it must have a limit point $x$. By continuity of the rotation function, $T^n(x)$ is a limit as well since taking $T$ of every term yields the same sequence (with only the first term removed). By induction, we have infinitely many distinct limit points $\{T^n(x)\}$.  That's all I could come up with! It was also easy to show that the orbit is infinite. I still don't seem to be able to get close to the required result however.
88078
Let $(X,T_1),\,(Y,T_2)$ be topological spaces and $H$ be a homeomorphism from $X \to Y$. If $C$ is a dense subset of $X$, is $H(C)$ necessarily dense in the $Y$?  I thank you in advance.
53536
These usually aren't too bad but I had difficulties thinking of what the set  $$\{z\in\mathbb{C}:|z+i|=2|z|\}$$ looks like in the complex plane. I got as far as  $$|z+i|=2|z|\Rightarrow \sqrt{(z+i)(\overline{z+i})}=2z\overline{z}\Rightarrow  \sqrt{z\overline{z}+i\overline{z}-iz+1}=2z\overline{z}\\ \Rightarrow z\overline{z}-i(z+ z\overline{z})+1=4z\overline{z}\Rightarrow 3z\overline{z}=1-i2\Re(z)$$  And rearranging a bit I guess I can see that $z\overline{z}+\frac{2i\Re(z)}{3}=1/3$. If there weren't that second term on the left, this would be clear, but I am not sure how to think about the dependence on the real part of $z$ here.   Having graphed in wolfram, I know this is just a circle of radius 2/3 centered at $i/3$, but I would appreciate help with the thought process.
121460
Let $E_n$ be $$E_n:=\{f\in C[0,1]\mid \text{exist } x_f\in[0,1] \text{ such that } |f(x)-f(x_f)|\leq n|x-x_f|,\, \forall x\in[0,1]\}.$$  How show that $E_n$ is nowhere dense, that is, $\mathrm{int}\  \overline{E_n} = \emptyset$?  Can someone help me?
33619
Well, after spending hours on this problem, I'm still stuck, so I thought I'd turn to you guys. The problem statement is as follows.  Let $f$ be a complex-valued function that is $C^1$ in the disk $|z - z_0| &lt; R$ for some $R&gt;0$, and let $C_r$ be the circle $|z - z_0| = r$. Show that $$\lim_{r\to 0} \frac{1}{r^2}\int_{C_{r}}f(z)dz = 2 \pi i\frac{\partial f}{\partial \bar{z}}(z_0).$$  I tried parametrizing the circle in the usual way and differentiating to see if I get something useful, but no luck. I just can't get the LHS to give me something to do with $\frac{\partial f}{\partial \bar{z}}$, so that I could then put the RHS under the integral and show that the difference between the two can be less than arbitrary $\epsilon &gt; 0$ by the continuity of partial derivatives.  Thanks in advance for your help!
63791
Consider the complex mapping defined by $$f_a(z)=\frac{z-a}{\bar az-1}$$ for some constant $a.$ I want to show that $f$ maps the unit circle to itself. I know that the unit circle is defined by $|z|=1,$ or equivalently $z=e^{i \theta}.$ This gives me $$f_a(z)=f_a(e^{i \theta})=\frac{e^{i \theta}-a}{\bar a e^{i \theta}-1}$$  I'm not sure how to proceed from here though.   Additional question: I also want to show that of $a$ lies inside the unit disk then $f$ maps the unit disk to itself. In order to answer this I have shown that $$|\bar a z-1|^2 - |z-a|^2 =(1-|a|^2)(1-|z|^2)&lt;1$$  How am I able to use this to deduce the result?
12572
Let $C$ denote the unit circle  centered at the origin in Complex Plane  What is the value of   $$ \frac{1}{2\pi i}\int_C |1+z+z^2 |dz,$$ where the integral is taken anti-clockwise along $C$?   0 1 2 3   What I have answered is 0 because it seems like $f(z)$ is analytic at 0 hence by Cauchy's Theorem.
27863
I am trying to show that the following space is not Hausdorff. Consider the topological space $S^1$, and let $r$ be an irrational number. Consider the action of $\mathbb{Z}$ on $S^1$ given by $$ S^1\times\mathbb{Z}\to S^1; (e^{ix}, n)\mapsto e^{i(x+2\pi n r)}. $$  Let $S^1/\mathbb{Z}$ denote the orbit space. I want to show that this space is not Hausdorff.   It was suggested to me that I try showing that any orbit under this action is dense in $S^1$, but I am getting stuck on proving that bit. But here is what I was thinking: we know that the topological group $\mathbb{R}/2\pi \mathbb{Z}$ is homeomorphic to $S^1$, as seen by the map $t\mapsto e^{it}$. Denote the following composition of maps $$ \mathbb{R}\to \mathbb{R}/2\pi \mathbb{Z}\simeq S^1\to S^1/\mathbb{Z} $$ by $\phi$. Then if $[e^{ix}]\in S^1/\mathbb{Z}$, it follows that  $$ \phi^{-1}([e^{ix}]) = \{x+2\pi(nr+m)\mid n, m\in\mathbb{Z}\}. $$ If I can show that this subset is dense in $\mathbb{R}$, then it follows that the set $[x]$ as a subset of $S^1$ is dense. This is where I am getting stuck, and it is not even clear to me that this is necessarily true.  Any hints or suggestions are appreciated. Thanks!
4852
I need to prove $\int_{0}^{\infty}\cos(t) t^{z-1}dt=\Gamma(z)\cos(\frac{\pi z}{2})$ for $0&lt;Re(z)&lt;1$. I tried $\cos t=\frac{e^{it}+e^{-it}}{2}$ and to integrate it along the contour from $\epsilon $ to R, then to $Ri$ via a quarter of a circle, then downward to $\epsilon i$, finally back to $\epsilon$ via a small quarter of a circle. However, I find that $e^{-it}t^{z-1} $ does not converge when $R\to \infty$, can anyone give a correction of my integration or offer another method to evaluate the integral?(would be better if using complex contours) thanks in advance.
122647
The problem is:  Suppose that $f(z)$ is continuous on a domain $D$ that contains the unit circle, and that $f(z)$ satisfies: $$\vert f(e^{i\theta})\vert \leq M \; \forall \theta \, \in [0, 2\pi )$$ and $$\bigg \vert \int _{\vert z \vert=1} f(z) \text{d}z \bigg \vert =2\pi M $$ show that there exists a $c\in \mathbb{C}$ such that $f(z)=c \bar{z}$ for all $\vert z\vert=1$.  I've tried lots of different approaches such as defining a function $g(z)=\frac{f(z)}{\bar{z}}$ and trying to show that $g'(z)=0$ so that $g$ is constant on the circle, or trying to use some bounding theorems for analytic functions to show that $g$ is constant but this requires $g$ to be analytic. Some other methods were using the Cauchy-Riemann equations but I couldn't find any reason why $g$ should be analytic.  Any help would be greatly appreciated, thanks!
30783
I'm having difficulty understanding how to prove that the primitive roots of unity are in fact dense on the unit circle. I have the following so far:  The unit circle can be written $D=\{x\in\mathbb{C}:|x|=1\}$. The set of primitive $m$-th roots of unity is $A_m=\{\zeta_k:\zeta_k^m=1,\zeta_k\text{ is primitive}\}$.  Hence, the set of all primitive roots $A$ is given by the union of $A_m$ over $m=1,2,3,\ldots$. But I can't seem to get started on how to prove that $A$ is dense in $D$.
7127
Clearly, $\mathbb Z$ is not dense in $\mathbb Q$ (and not dense in $\mathbb C$).  But why is the ring of algebraic integers $\overline{\mathbb Z}$ dense in $\overline{\mathbb Q}$?  In particular, if this is so, it is dense in $\mathbb C$.
29631
There are three parts of the question, the first two which I proved.   $a.$ Proving $f(z)$ is entire analytic.  $b.$ $f(z+1) = f(z)$ and $f(z+i) = e^{\pi}e^{-2\pi iz}f(z)$  $c.$ Inside the unit square in the 1st quadrant, prove that $f(z)$ has a unique $z_0$ such that $f(z_0)=0$  I need some help regarding the $c.$ part of the question. I don't understand how to apply the argument principle here (which is a hint). It is also given that $f(z)$ does not vanish on the boundary of the square.  What I could do:  $$f(0) = f(1) = 1+2\sum_{n=1}^{\infty}e^{-\pi n^2}$$  and   $$f(i) = f(1+i) = e^{\pi}\big(1+2\sum_{n=1}^{\infty}e^{-\pi n^2}\big)$$  Using the hint given below by a user: I end up with $$\dfrac{1}{2\pi i}log f(z)$$ and the integral evaluates to -1/2 and 1/2 on the two vertical sides and 0 on the horizontal lines. Resulting in the closed integral to 0. Any ideas where I might be going wrong?  Any idea how to proceed?
27415
Is $\{n \sin n | n \in \mathbb{N}\}$ dense on the real line?  If so, is $\{n^p \sin n | n \in \mathbb{N}\}$ dense for all $p&gt;0$?  This seems much harder than showing that $\sin n$ is dense on [-1,1], which is easy to show.  EDIT: This seems a bit harder than the following related problem, which might give some insight:  When is $\{n^p [ \sqrt{2} n ] | n \in \mathbb{N}\}$ dense on the real line, where $[\cdot]$ is the fractional part of the expression?  I am thinking that there should be some probabilistic argument for these things.  EDIT 2:  Ok, so plotting a histogram over $n \sin n$ is similar to plotting $n \sin(2\pi X)$ where $X$ is a uniform distribution on $[-1,1].$ This is not surprising, since $n$ mod $2\pi$ is distributed uniformly on  $[0,2\pi].$  Now, the pdf of $\sin(2\pi X)$ is given by $f(x)=\frac{2}{\pi \sqrt{1-x^2}}$ in $(-1,1)$ and 0 outside this set.  The pdf for $n \sin(2\pi X)$ is $g_n(x)=\sum_{k=1}^n \frac{1}{nk} f(x/k)$ so the limit density is what we get when $n \rightarrow \infty.$ (This integrates to 1 over the real line).  Now, it should be straightforward to show that for any interval $[a,b],$ $\int_a^b g_n(x) dx \rightarrow 0$ as $n \rightarrow \infty.$  Thus, the series $g_n$ is "too flat" to be able to accumulate positive probability anywhere. (The gaussian distribution on the other hand, has positive integral on every interval).
23507
I just came across the problem of what happens in the boundary cases of .  More concisely, for what $\theta$ will the following sum converge? $$\sum_{n=1}^{\infty} \frac{e^{ni\theta}}{n}$$ It is well known that this sum will diverge for $\theta=0$ and will converge for $\theta=\pi$. In fact, for any odd multiple of an even root of unity, since we have symmetry, for each line through the origin of the complex plane, we can use the alternating series theorem to show that they converge.  Another idea I had was to split  the sum into imaginary and real parts, and if we can prove each of these converges, then we are all set. So now we must prove that both of the following converge $$\sum_{n=1}^{\infty}\frac{\cos(n\theta)}{n}\space, \space \sum_{n=1}^{\infty}\frac{\sin(n\theta)}{n}$$ Now these can be shown to converge easily by Dirichlet's test for any rational multiple of pi, but what about irrational multiples of pi? I know it has been shown to converge with $\theta=1$, but I don't see how to generalize this.
8142
I'm looking for an understandable proof of this theorem, and also a complex one involving beautiful math techniques such as analytic number theory, or something else. I hope you can help me on that. Thank you very much
94405
Can somebody give rigorous proof that the circle cannot be embedded into the real line? Obviously, if so it must be necessarily continuous and I guess this suffices in this case since there is no continuous injection by intuition...