Is there an entire not constant function with bounded real part?  I've seen a solution which involves the exponential and Liouville's theorem.  I wanted to see a more direct proof. If $f$ is an entire function and we have a continuously differentiable function $\gamma:[0,\infty)\rightarrow\mathbb C$ and we know that $\Im (f\circ\gamma)(t)\rightarrow\infty$ (for $t\rightarrow\infty$), can we construct an analogous curve for the real part?  I'm having trouble to get an intuition for how the real part of a holomorphic function looks like in relation to its imaginary part. The only thing I can visualize is that the gradient of the two are orthogonal, but I kind of struggle what that means for the potential of those two gradients, i.e. the real and imaginary part.
Greets  This is exercise 15.d chapter 3 of Stein &amp; Shakarchi's "Complex Analysis", they hint: "Use the maximum modulus principle", but I didn't see how to do the exercise with this hint rightaway, instead I knew how to do it with the Casorati-Weiestrass Theorem, here is my answer:  Define $g(z)=f(1/z)$ for $z\neq{0}$,then by the hypothesis we must have that for any $\epsilon&gt;0$ $g(D_{\epsilon&gt;0}(0)-\{0\})$ is not dense in $\mathbb{C}$, then the singularity at $0$ of $g$ is not essential, this implies $f$ must be a polynomial, but if $f$ is a non-constant polynomial, it is easy to see that its real part must be unbounded, so $f$ must be constant.  I would like to know an answer with the the maximum modulus principle.  Thanks
Enter from password field doesn't submit SE login form On  page for a site where I don't yet have an account, if I select the "Log in using Stack Exchange" box, when I hit enter from the password form, nothing happens.  Minor, but... disconnected phone while installing ROM I recently tried to install a custom ROM in my samsung j6+ but halfway through it my PC turned off. Then the problem starts i decided to reboot my phone but it shows this error &quot;No command&quot; &quot; #Reboot recovery cause is [[check_bootmode]RebootRecoveryWithKey]# Block-based OTA Supported API: 3 # MANUAL MODE v1.0.0# E:Failed to mount / cache: Invalid argument  The error message keeps appearing for just a fraction of a second then disappear and appear again I tried to go to recovery mode but no luck i cant get to it I drained the battery to turn off the device then go download mode but after get to it, it also keep restarting help me please. thank you Would the One Ring have stayed hidden if Bilbo hadn't turned up? Why is the charge naming convention wrong? I recently came to know about the Conventional Current vs. Electron Flow issue. Doing some search I found that the reason for this is that Benjamin Franklin made a mistake when naming positive and negative charges. There is even this little comic about that   My question is, how can a naming convention be wrong? Perhaps I don't understand what is the actual mistake here, I don't know. But I don't see how a naming convention could be wrong or right. There is no right and wrong about that. It could've been any other name, like charge A and charge B.  I'll appreciate anyone can help me understand what is wrong in all this. Captchas not displaying in SO for me When I get a captcha on Stack Overflow, I don't see the image or the input box with Firefox. I originally had the "HTTPS Anywhere" extension installed but still have the same issue when it's disabled (This caused issues previously when Google's SSL certs were expired for an old domain IIRC).  The requests to Google seem to be never-ending; browsing to the URLs that are still loading results in some pages of JavaScript, but no errors as you'd hope.  The screenshot was taken after the page had been open for more than 3 minutes, and I'm not sure why the GETs are still running. The request popup visible in Firebug is for "GET challenge?k=...."  When browsing to the same captcha page in IE, there are no issues.  I'm aware this could be an issue with my Firefox configuration, as opposed to the site, but would appreciate it if someone could verify if it's just me, and suggest a fix.  EDIT   While visiting the Recaptcha site, I can see a Captcha with no problems If I right-click the problem URLs in FF and Open in new tab, the pages load fine. From the fact that FB omits IP Addresses for the problem ones, I assume that it's not even opening a connection (nor perhaps performing a DNS lookup) Other sites that use Recaptcha seem to work fine (I picked 3 at random from Google) Where does the word "dog" come from? Plurals of acronyms, letters, numbers — use an apostrophe or not? When I was in high school back in the 1970s, I was taught that to make a plural of an acronym, a letter, or a number, one should add an apostrophe and "s". Like I would have written this sentence, "... back in the 1970's ..." I would write "one CD, two CD's". Etc. I followed this rule faithfully for years until a co-worker told me it was wrong. Now I can't find any source that agrees with what I was taught.  Is this a rule that has changed over time? Was the convention in the 70s that one should use an apostrophe but this has changed and now we do not? Or were my high school English teachers just confused? How do I make a 2D game resolution independent? I'm working on a simple 2D game. I've finished the mobile phone version.  However, my boss wants the game to works on his RT. I'm doing the "conversion" but my buttons are in the wrong places, because I hard-coded the screen size, like this:   texture = game.Content.Load&lt;Texture2D&gt;("StartScreen");  mainFrame = new Rectangle(0, 0, game.GraphicsDevice.Viewport.Width, game.GraphicsDevice.Viewport.Height);   // button definitions  startbrect = new Rectangle(300, 258, 88, 88);  startbtext = game.Content.Load&lt;Texture2D&gt;("bplay");   In this example, mainframe is fine, but startbrect isn't, because I defined the size to match a Windows phone's screen. How can I handle responsive design when all Windows 8 phones' screens are different? Is there a formula or macro to calculate each time the good size? Let $k$ be a finite field of order $q$ in characteristic $p$, let $n$ be a positive integer not divisible by $p$, and let $K$ be the splitting field of $X^n-1$ over $k$. Prove that $[K:k]$ equals the smallest positive integer $d$ such that $n\mid q^d-1$.  My approach: It's easy to show that if $k$ is a finite field of order $q$ with $\text{char} \ k=p$ then $k=\mathbb{F}_{q}$ with $q=p^m$ with $m\geq 1$. If $K$ is the splitting field of $X^n-1$ over $k$ then $K=\mathbb{F}_{q^r}$ for some suitable $r\geq 1$.  Let $\alpha \in K=\mathbb{F}_{q^r}$ be a root of $X^n-1$ then $\alpha^n-1=0$ and $\alpha\neq 0$.   Thus, $\alpha^n=1$ and $\alpha \in \mathbb{F}^{\times}_{q^r}$ where $\mathbb{F}^{\times}_{q^r}$ is the cyclic group of order $q^r-1$.  But I have some difficulties:  1) From $\alpha^n=1$ it follows that $o(\alpha)\mid n$. How to show that $o(\alpha)=n$? I was trying to show it using that $(n,p)=1$ but I failed.  Remark: Because if we can show that $o(\alpha)=n$ then the desired result follows immediately from cyclicity of multiplicative group of field $\mathbb{F}_{q^r}$.  I have spent one day in order to come up with this approach. So please do not duplicate this question and can anyone show how to answer my question, please.  Detailed explanation would be great! Simple prime number generator in Python I found an argument online that the two fields are not isomorphic but I can't make sense of the argument. You find it .  These are the things I'm confused about when it comes to the argument.     It says that the field $\mathbb Q[x]/(x^2+1)$ contains a root of the polynomial $x^2+1$. I suppose this is the congruence class $[x]$? But this is not the same as the zero element of the field $\mathbb Q[x]/(x^2+1)$, right?   It says that every isomorfism takes $0$ to $0$. I interpret this as the zero-element in $\mathbb Q[x]/(x^2+1)$ (which I interpret to be $[0]$) is taken to the zero-element in $\mathbb Q[x]/(x^2+2)$ denoted $[0]_2$. But in the argument it says that this means that if $\mathbb Q[x]/(x^2+1)$ has a root of the polynomial, then also $\mathbb Q[x]/(x^2+2)$ must have a root of this same polynomial. Why is that?     Greatful for any help! How can I use QGIS to create a metro style map We consider $g : (t,u,v)\in \mathbb{R}^3 \mapsto (t^2,u^2,v^2,\sqrt{2}uv, \sqrt{2}tv,\sqrt{2}tu)\in\mathbb{R}^6$. I have to prove that $g(\mathbb{S}^2)$ is a submanifold of $\mathbb{R}^6$.  $dg_{(t,u,v)}=\begin{pmatrix}2t&amp;0&amp;0\\  0&amp;2u&amp;0\\  0&amp;0&amp;2v\\  0&amp;\sqrt{2}v&amp;\sqrt{2}u\\ \sqrt{2}v&amp;0&amp;\sqrt{2}t\\ \sqrt{2}u&amp;\sqrt{2}t&amp;0\\ \end{pmatrix} $  So it proves that $g$ is an immersion (the rank is $3$ clearly) for $(\mathbb{R}^3-\{(0,0,0)\})$.  Now how to prove that $\mathbb{S}^2$ is homeomorphic to $g(\mathbb{S}^2)$ ? I must prove that $g$ is injective  ?  Thanks in advance ! Using Iterator with ModelBuilder for clipping and splitting in ArcMap? Does the artificer's Enhanced Weapon stack on top of hexblade's Improved Pact Blade? iPhone asks 'Trust This Computer' again (I am on macOS 10.14.6 and iOS 12.4)   I am certain that I have trusted my Mac before. Without the latter trust, I can also access the phone in iTunes.   I have tried reset "Location and Privacy" on my phone. Then   When connecting the iPhone with iTunes running, it asks for the trust and I trust it.  Reboot my Mac and the phone asks for another trust.   Is there some known malware mimics system behavior and steal data from an iPhone?  (I do have installed some pir*ted software recently. That's my fault.) I cannot submit my answer to    I am aware that my reputation is low. Also, spam posting is discouraged.  My answer includes a compilation of Android graphing libraries, with links, so fellow hackers do not have to Google each and every one of them.  I believe this answer is superior to the current answers, because it is updated to 03/2012 and covers more than 90% of all Android charting libraries.  How can I post the answer, or how can I increase my reputation without being able to post quality answers? The old school websites included just a bunch of links to other places, I am an old man!  Alternative solution would be to blog it, and just post a link to my blog, which is not optimal. Suppose the terms of the sequence of real numbers $\{a_n\}$ satisfy $|a_{n+1} - a_n| \le 2^{-n}$ for all $n$. Prove that $\{a_n\}$ is Cauchy.   My Work  So by the definition of a Cauchy sequence, for all $\varepsilon &gt; 0$ $\exists N$ so that for $n,m \ge N$ we have $|a_m - a_n| \le \varepsilon$. However, questions like  make me understand that the $2^{-n}$ condition is necessary for this to be a true statement.   So I am wondering how to appeal to the Cauchy definition for this proof. Do I prove that every convergent sequence is therefore Cauchy, and then try to prove convergence? I'm having some problems with listings and UTF-8 in my document. Maybe someone can help me? Some characters work, like é and ó, but á and others appear at the beginning of words...  \documentclass[12pt,a4paper]{scrbook} \KOMAoptions{twoside=false,open=any,chapterprefix=on,parskip=full,fontsize=14pt}  \usepackage[portuguese]{babel} \usepackage[utf8]{inputenc} \usepackage[T1]{fontenc} \usepackage{listingsutf8}  \usepackage{inconsolata} \lstset{     language=bash, %% Troque para PHP, C, Java, etc... bash é o padrão     basicstyle=\ttfamily\small,     numberstyle=\footnotesize,     numbers=left,     backgroundcolor=\color{gray!10},     frame=single,     tabsize=2,     rulecolor=\color{black!30},     title=\lstname,     escapeinside={\%*}{*)},     breaklines=true,     breakatwhitespace=true,     framextopmargin=2pt,     framexbottommargin=2pt,     extendedchars=false,     inputencoding=utf8 }  \begin{document} \begin{lstlisting} &lt;?php  echo 'Olá mundo!'; print 'Olá mundo!'; \end{lstlisting}  \end{document} \end{lstlisting} How to handle OutOfMemoryError in Java?