Suppose $a \in \mathbb{C}$, $|a| < 1$, and $f(z) = \dfrac{z - a}{1 - \overline{a}z}$. How to prove dependence of $|f(z)|$ on $|z|$?
Is there anything special with complex fraction $\left|\frac{z-a}{1-\bar{a}{z}}\right|$?
Can a regular salaried (w2) employee opt to pay quarterly taxes? As a consultant, I got used to paying quarterly taxes. Now that I'm a regular w2 salaried employee, I'm subject to normal withholding on each pay statement. I would like to keep making exact quarterly payments instead of having a (generally incorrect) amount held back each paycheck.   I haven't been able to find any information on the 1040-es that would make me think I'm ineligible to do so, but I suspect that it may be the case. I'd have to stop paycheck withholdings by updating my w4 to either have a high number of exemptions or mark myself as exempt from taxes, which I assume is where the rub lies.   So, as a regular salaryman, is it possible for me to pay taxes exclusively through quarterly payments in lieu of traditional withholdings? MyClass[] array; List&lt;MyClass&gt; list;   What are the scenarios when one is preferable over the other? And why? My question is from Apostol's Vol. 1 One-variable calculus with introduction to linear algebra textbook.  Page 57. Exercise 12. Show that the following are polynomials by converting them to the form $\sum_{k=0}^{m}a_kx^k$ for a suitable $m$. In each case $n$ is a positive integer. $a)$ $(1+x)^{2n}.$ $b)$ $\frac{1-x^{n+1}}{1-x}, x\not=1.$ $c)$ $\prod_{k=0}^{n}(1+x^{2^k}).$  The attempt at a solution: a) part of the problem is pretty easy I guess, it is example of binomial theorem, so the answer would be $(1+x)^{2n}=\sum_{k=0}^{2n}(^{2n}_{k})x^k.$ Answer to part b) would be the following: $$\frac{1-x^{n+1}}{1-x}=\frac{(1-x)(1+x+x^2+\cdots+x^n)}{1-x}=1+x+x^2+\cdots+x^n=‌​\sum_{k=0}^{n}x^k.$$thanks to @DiegoMath's hint. As for part c), we have $$\prod_{k=0}^{n}(1+x^{2^k})=(1+x)(1+x^2)(1+x^4)(1+x^8)\cdots(1+x^{2^n})$$ and I have trouble &quot;converting&quot; this to sum which would be of a form of polynomial. Show that  $A \setminus ( B \setminus C ) \equiv ( A \setminus B) \cup ( A \cap C )$  This is an exercise I was trying to do ( not homework ) and I got stuck as follows:  Working from $A \setminus ( B \setminus C )$:  $x \in A \wedge x \notin ( B \setminus C )$ $x \in A \wedge ((x \in B \wedge x \in C) \vee (x \notin B \wedge x \in C) \vee (x \notin B \wedge x \notin C) )$  At this point I'm not sure how to proceed.  Working from the other side $ ( A \setminus B) \cup ( A \cap C ) $:  $ x \in ( A \setminus B) \cup ( A \cap C ) $  $ x \in ( A \setminus B) \vee x \in ( A \cap C ) $  $ ( x \in A \wedge x \notin B ) \vee ( x \in A \wedge x \notin C ) $  $ x \in A \wedge (x \notin B \vee x \notin C ) $  $ x \in A \wedge x \notin ( B \cap C ) $  $ A \setminus ( B \cap C ) $  I could also use some formating tips like how display logical not and how to get the lines to display properly without blank lines inbetween. :-) Math Induction Proof: $(1+\frac1n)^n < n$ So I have to prove:      For each natural number greater than or equal to 3,   $$(1+\frac1n)^n&lt;n$$   My work: Basis step: $n=3$ $$\left(1+\frac13\right)^3&lt;3$$ $$\left(\frac43\right)^3&lt;3$$ $$\left(\frac{64}{27}\right)&lt;3$$ which is true.  Now the inductive step, assume $P(k)=\left(1+\frac1k\right)^k&lt;k$ to be true and prove $P(k+1)=\left(1+\frac1{k+1}\right)^{k+1}&lt;k+1$.   This is where I am stuck because usually you add or multiply by $k+1$ or some similar term. How do I delete data for one game in the iPad Game Center? I have currently a strange problem on debian (wheezy/amd64).  I have created a chroot to install a server (i can't give any more detail about it, sorry). Let's call its path /chr_path/. To make things easy, I have initialized this chroot with a debootstrap (also wheezy/amd64).  All seemed to work well inside the chroot but when I started the installer script of my server I got : zsh: Not found /some_path/perl (the installer includes a perl binary for some reasons)  Naturally, I checked the /some_path/ location and I found the "perl" binary. file in chroot environment returns :  /some_path/perl ELF 32-bit LSB executable, Intel 80386, version 1 (SYSV), dynamically linked (uses shared libs), for GNU/Linux 2.2.5, not stripped   The file exists, seems ok, has correct rights. I can use file, ls, vim on it but as soon as I try to execute it - ./perl for example - I get : zsh: Not found ./perl.  This situation is quite understandable for me. Moreover :   I can execute other basic binaries (/bin/ls,...) in the chroot without getting errors I have the same problems for other binaries that came with the project When I try to execute the binary from the main root (/chr_path/some_path/perl), it works. I have tried to put one of the binaries with a copy of my ls. I checked that the access rights were the same but this didn't change anything (one was working, and the other wasn't) How to log commands within a "sudo su -"? If I:   [user@notebook ~] sudo echo 123456uu 123456uu [user@notebook ~]    Then I can see that in the logs:  [root@notebook /var/log] grep 123456uu * auth.log:Jan  9 17:01:51 notebook sudo: user : TTY=pts/3 ; PWD=/home/user ; USER=root ; COMMAND=/bin/echo 123456uu [root@notebook /var/log]    but if I:   [user@notebook ~] sudo su - [root@notebook ~] echo 1234567zz 1234567zz [root@notebook ~]    I cannot see it in the logs:   [root@notebook /var/log] grep 1234567zz * [root@notebook /var/log] echo $? 1 [root@notebook /var/log]    My question: How can I turn the logging on for the commands within the "sudo su -"?  OS is an Ubuntu 12.04 but the question is in general.   UPDATE#1:   [user@notebook ~] sudo su - [sudo] password for user:  [root@notebook ~] echo zizizi zizizi [root@notebook ~] cd /var/log [root@notebook /var/log] grep -iIR 'zizizi' * [root@notebook /var/log] grep 'COMMAND=/bin/su -' * auth.log:Jan 10 15:42:42 notebook sudo: user : TTY=pts/1 ; PWD=/home/user ; USER=root ; COMMAND=/bin/su - [root@notebook /var/log] Show that the congruence $x^{p-1} \equiv 1 \pmod {p^2}$ has precisely $p-1$ solutions modulo $p^2$ Let $p$ be a prime number. Show that the congruence $x^{p-1} \equiv 1 \pmod {p^2}$ has precisely $p-1$ solutions modulo $p^2$  I know the statment above is true for modulo $p$, but how is this imply to modulo $p^2$ Convert .jar to .deb package Is it possible to split micro USB to USB cable so two cables (data + and data -) go to the PC with USB and two other power cables go to power brick? Connectedness in the Quotient Space Determine value of the limit:  $$L=\lim_{n\to \infty}\sqrt[n]{1^2+2^2+\cdots+n^2}$$  My try:  $$1^2+2^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}$$  Hence: $$L=\lim_{n\to \infty} \sqrt[n]{\frac{n(n+1)(2n+1)}{6}}=?$$  But, come here, i do not know how, because please help me. How do I go about tracking down IO usage in Linux and other *nix OS'? Occasionally, something on my system starts hammering the disk. I would like to log data that would let me know about I/O usage per process, then I could go back to it later and see what keeps doing this. How should I go about solving this problem? Bonus points for solving this for arbitrary block devices. Stack Exchange app iPhone X optimization The Stack Exchange app doesn’t look very good on Apple’s brand new flagship phone yet.    Also the app can’t identify the device model either. (Says Unknown (iPhone 10,3))   App Version: 1.6.6.2 Device: iPhone X (iPhone10,3) OS Version: Version 11.1 (Build 15B93) Show that product of primes, $\prod_{k=1}^{\pi(n)} p_k < 4^n$ This an interesting problem my friend has been working on for a while now (I just saw it an hour ago, but could not come up with anything substantial besides some PMI attempts).  Here's the full problem:  Let $x_{1}, x_{2}, x_{3}, \cdots x_{y}$ be all the primes that are less than a given $n&gt;1,n \in \mathbb{N}$.   Prove that $$x_{1}x_{2}x_{3}\cdots x_{y} &lt; 4^n$$  Any ideas very much appreciated! How to create a table with long text and which packages use Random bits of my chat die and come back to life during the day How are the age restrictions followed by ICC for under 19 cricket tournaments? I have some doubts under this topic,   Can a player aged 19 play under 19 international cricket match? (Since 19 is not under 19) Consider a player having birthday during the series/tournament. How the age will be calculated for him? Why can't I buy the Windhelm house (Hjerim)? I've finished the Stormcloak storyline, and now cannot become a Thane of Windhelm without buying Hjerim. The Steward keeps saying something has happened and he can't sell it right now.  What the heck am I missing? I've been doing every random little quest from people in town I can find, hoping to unlock the house. I'm well over the 5/5 requirement of helping people.  Any ideas?