\documentclass[12pt]{amsart}
\usepackage{amsfonts}
\setlength{\topmargin}{0in}
\setlength{\textheight}{10in}
\setlength{\oddsidemargin}{0in}
\setlength{\evensidemargin}{0in}
\setlength{\textwidth}{6.5in}
\parskip0.4em
\thispagestyle{empty}
\def\rr{\mathbb{R}}
\def\Z{\mathbb{Z}}
\def\Q{\mathbb{Q}}
\def\C{\mathbb{C}}
\def\G{\mathbb{G}}
\renewcommand{\phi}{\varphi}

\begin{document}
\begin{center}
{\bf Math 113 Homework 1, due 2/5/2019}
\end{center}
\medskip

{\bf 1.} Write an addition table for the integers modulo $3$ (it should have nine entries). 

{\bf 2.}
(a)
Multiply the following complex numbers by $i$ and by $-i$: $$2+i, -i, -\pi + e\cdot i, -\sqrt{2}-2i.$$

\

(b) If $z = a + b i$ is a complex number, its \emph{complex conjugate} is the number $$\bar{z}: = a - bi$$ with the imaginary part negated. Show that $z\cdot \bar{z} = a^2 + b^2$. (This implies in particular that it is a real, positive number.) Show (this should be easy) that $iz\cdot \bar{iz}$ gives the same answer as $z \cdot \bar{z}$.

\

(c) Find $8$ distinct Gaussian numbers $z$ such that $z\cdot \bar{z} = 5$
(remember that $z\in \C$ is a Gaussian number if $z = a+bi$ for $a,b \in \Z$). Note: the shorthand notation $\pm$ will be useful here; for example, $\pm 1 \pm i$ represents shorthand for a set of four different complex numbers!

\

(d) Find $12$ distinct Gaussian numbers $z$ such that $z\cdot \bar{z} = 25$. Hint: multiply together pairs of numbers in part (c).

\

(e) Show that there are no Gaussian numbers $z$ such that $z\cdot \bar{z} = 3$ (hint: squares of nonzero integers are always $\ge 1$).

\

{\bf 3.} 
(a) Check that multiplication of Gaussian numbers modulo $2+i$ is well-defined, i.e. if $\alpha\sim \alpha'$ are Gaussian numbers which are equivalent modulo $2+i$ (i.e.\ their difference is a Gaussian number divisible by $2+i$) and $\beta\sim \beta'$ are two other Gaussian numbers which are equivalent to each other then $\alpha\cdot \alpha' \sim \beta\cdot \beta'$ and $\alpha + \alpha' \sim \beta + \beta'$. Hint: write $\alpha' = \alpha + \delta\cdot (2+i),$ etc.

{\bf Remark:} of course the specificity of working modulo $2+i$ is not important here: we're using the example of $2+i$ to check that modular arithmetic works for the Gaussian numbers.

\

(b) Remember that each Gaussian number is equivalent (modulo $2+i$) to one of the five elements $\{0, 1, i, -1, -i\}$. Write down an addition table for the five classes of elements, namely $[0]$ (any element equivalent to $0$), $[i]$ (any element equivalent to $i$) etc. So for example, $[0] + [i] = [i]$ (here the calculation works on the nose), but $[1] + [i] = [-1],$ since $1+i$ is not on the liest but $1+i \sim -1$ (their difference is $2+i$), and $-1$ is. This table should have 25 entries (though half of them are immediate because of commutativity).

\

(c) Similarly, write down a multiplication table. (This will be surprisingly easy: this is a special nice property of residues modulo $2+i$.)

\ 

{\bf 4.} (a) Now write down the addition table $\Z/5,$ the integers modulo $5$. Can you find an isomorphism $$f:(\Z/5, +)\to (\G/(2+i), +)$$ of binary structures from the integers modulo $5$ with addition to the Gaussian integers modulo $2+i$ (from the previous problem), also with addition? Can you find another, different isomorphism? (To specify an isomorphism, write down a class of Gaussian integers modulo $2+i$ where each element of $\Z/5$ goes, and check a couple of nontrivial cases of the homomorphism property to convince yourself that this is actually an isomorphism.) Hint: if you first pick where $[1]\in \Z/5$ goes, you also know where $[2] = [1]+[1]$ goes by the homomorphism property, and then you know where $[3] = [2] + [1]$ goes, etc.

\

{\bf 5.} View $2$ as a Gaussian number. Then if $\alpha = a+bi$ is another Gaussian number, then $2\alpha = 2a + 2bi,$ so a Gaussian number is divisible by $2$ if both of its components are even, and two Gaussian numbers are equal modulo $2$ if their difference has even components. This gives four equivalence classes of Gaussian numbers modulo $2$, namely:
$$\text{ even } + \text{ even }\cdot i, \text{ odd } + \text{ even }\cdot i, \text{ even } + \text{ odd }\cdot 1, \text{ and odd } + \text{ odd }\cdot i.$$ We pick a ``representative'' Gaussian number in each class and write these in shorthand (in the same order) as $\{[0], [1], [i], [1+i]\}.$ Now write addition and multiplication tables for the Gaussian numbers modulo $2$. 

\

{\bf 6.} {\bf Challenge problem: you can do this one instead of any two of the others, or you can do it as extra credit for an extra 10\% on this HW.}

(a) Show that any complex number is a distance of at most $\frac{\sqrt{2}}{2}$ away from a Gaussian number (hint: $\frac{\sqrt{2}}{2}$ is the distance from a vertex of the unit square to its center). 

\

(b) Show that if $\alpha$ is a nonzero complex number, then any complex number $z$ is at most $\frac{\sqrt{2}|\alpha|}{2}$ away from a multiple of $\alpha$ (hint: if $z$ is distance $d$ from $\lambda\cdot \alpha$ then $\frac{z}{\alpha}$ is distance $\frac{d}{|\alpha|}$ away from $\lambda$).

\

(c) Deduce that if $\alpha = a + bi$ is a nonzero Gaussian number then any other Gaussian number $z\in \G$ is equivalent modulo $\alpha$ to a Gaussian number of magnitude $<|\alpha|$ (remember that $|\alpha| = \sqrt{a^2 + b^2}$). This is a version of ``division with remainder'' for Gaussian numbers.

\

(d) This means that all ``colors'', or types of residue modulo $\alpha$ are contained in the interior of the circle of radius $|\alpha|$ (in fact, we've seen they are contained in the closed circle of radius $\frac{\sqrt{2}}{2}\alpha$). Give an example, however, where two such residues are the same (i.e.\ we are ``overcounting'').

{\bf Remark: from part (c) above, you can deduce that there are less than $\pi r^2$ residue classes modulo $\alpha,$ where $r^2 = |\alpha|^2 = a^2 + b^2.$ This is nice, because it implies that there are finitely many possibilities for the residue, which was not obvious a priori. Part (d) shows that this estimate will tend to overcount. In fact this estimate is off by a constant. There is a beautiful formula for the total number of residues: it is exactly $a^2 + b^2$. While you don't have to do this, I would encourage you to check this formula when $\alpha = a + 0i$ is real and positive.}

\end{document}