\newlec

\hdc
What is the value of $\left|\left(\overline{-\sqrt{3}+i}\right)^5\right|$ ?
 
\begin{enumerate}[A)]
\item $16i$
\item $27$
\item $32$
\item $-45$
\item $64$
\end{enumerate}
\vskip3em
\soln
\ic{Emphasize there are lots of ways to get the solution.}
\begin{align*}
\left|\left(\overline{-\sqrt{3}+i}\right)^5\right| &= \left|\left(-\sqrt{3}-i\right)^5\right|\\
&= \left|\left(-\sqrt{3}-i\right)\right| ^5&&\text{PM}\\
&= \sqrt{(-\sqrt{3})^{2} + (-1)^{2}}^{5}\\
&= \sqrt{4}^{5} \\
&= 32
\end{align*}
\ic{This is the 7-10 minute mark depending on how many ways you find the above answer}
\newpage

\textbf{Polynomials}
For us, a field will mean to include $\mathbb{Q}, \mathbb{R}, \mathbb{C}, \mathbb{Z}_{p}$ where $p$ is a prime number. A ring will include the aforementioned fields as well as $\mathbb{Z}$ and $\mathbb{Z}_{m}$ for any $m\in\mathbb{N}$.
\vskip1em
\defn A polynomial in $x$ over a ring $R$ is an expression of the form
\[
a_{n}x^{n} + a_{n-1}x^{n-1} + ... + a_{1}x + a_{0}
\]
where $a_{0}, a_{1},... a_{n} \in R$ and $n \geq 0$ is an integer. Denote the set (actually a ring) of all polynomials over $R$ by $R[x]$. 
\ic{We will predominately use fields in the above definition. Some of the theorems we do will only work in the case of fields. For simplicity I will state all the theorems with fields to match the textbook though in many cases, a ring is all you need.}
\vskip1em
\ex \begin{enumerate}
\item $(2\pi +i)z^{3} - \sqrt{7}z + \tfrac{55}{4}i \in \mathbb{C}[z]$.
\item $[5]x^{2} + [3]x + [1] \in \mathbb{Z}_{7}[x]$. We usually write this as $5x^{2} + 3x + 1 \in \mathbb{Z}_{7}[x]$.
\item $x^{2} + \tfrac{1}{x}$ is \textit{not} a polynomial.
\item $x + \sqrt{x}$ is \textit{not} a polynomial.
\item $1 + x + x^{2} + ...$ is \textit{not} a polynomial.
\end{enumerate}
\defn \begin{enumerate}
\item The coefficient of $a_{n}x^{n}$ is $a_{n}$
\item A term of a polynomial is any $a_{i}x^{i}$
\item The degree of a polynomial is $n$ provided $a_{n}x^{n}$ is the term with the largest exponent on $x$ and nonzero coefficient.
\item $0$ is the zero polynomial (all coefficients are 0). The degree of the zero polynomial is undefined (some books say it is negative infinity for reasons we will see later)
\item A root of a polynomial $p(x) \in R[x]$ is a value $a \in R$ such that $p(a) = 0$.
\item If the degree of a polynomial is
\begin{itemize}
\item $1$, then the polynomial is linear.
\item $2$, then the polynomial is quadratic.
\item $3$, then the polynomial is cubic.
\end{itemize} 
\item $\mathbb{C}[x]$ are the complex polynomials, $\mathbb{R}[x]$ are the real polynomials, $\mathbb{Q}[x]$ are the rational polynomials, $\mathbb{Z}[x]$ are the integral polynomials.
\item Let 
\begin{align*}
f(x) = a_{n}x^{n} + a_{n-1}x^{n-1} + ... + a_{1}x + a_{0} \quad\quad \text{ and } \quad\quad  g(x) = b_{n}x^{n} + b_{n-1}x^{n-1} + ... + b_{1}x + b_{0}
\end{align*}
be polynomials over $R[x]$. Then $f(x) = g(x)$ if and only if $a_{i} = b_{i}$ for all $i \in \{0,1,...,n\}$.
\item $x$ is an indeterminate (or a variable). It has no meaning on it's own but can be replaced by a value whenever it makes sense to do so.
\item Operations on polynomials: Addition, Subtraction, Multiplication (See next page)
\end{enumerate}
\ic{This is probably the 25-30 minute mark. The lecture is a bit dry but we need to be on the same page.}
\newpage

\hdc
Simplify $(x^{5} + x^{2} + 1)(x+1) + (x^{3} + x + 1)$ in $\mathbb{Z}_{2}[x]$
\vskip3em
\soln 
\begin{align*}
(x^{5} + x^{2} + 1)(x+1) + (x^{3} + x + 1) &= x^{6} + x^{5} + x^{3} + x^{2} + x + 1 + x^{3} + x + 1\\
&= x^{6} + x^{5} + 2x^{3} + x^{2} + 2x + 2\\
&= x^{6} + x^{5} + x^{2}
\end{align*}

\newpage

\ex Prove that $(ax +b)(x^{2} + x + 1)$ over $\mathbb{R}$ is the zero polynomial if and only if $a=b=0$.
\vskip1em
\pf
 Expanding gives
 \begin{align*}
(ax+ b)(x^{2} +x + 1) = ax^{3} + (a+b)x^{2} + (a+b)x + b.
\end{align*}
This is the zero polynomial if and only if $a=0$, $a+b=0$ and $b=0$ which holds if and only if $a=b=0$. \qed
\ic{This is the 40 minute mark}
\thm (Division Algorithm for Polynomials (DAP)) Let $\mathbb{F}$ be a field. If $f(x), g(x) \in \mathbb{F}[x]$ and $g(x) \neq 0$ then there exists unique polynomials $q(x)$ and $r(x)$ in $\mathbb{F}[x]$ such that 
\[
f(x) = q(x)g(x) + r(x)
\]
with $r(x) = 0$ or $\deg(r(x)) < \deg(g(x))$.
\vskip1em
\pf Exercise (or extra reading). \qed
\vskip1em
\note
\begin{enumerate}
\item $q(x)$ is the quotient.
\item $r(x)$ is the remainder.
\item If $r(x)= 0$, then $g(x)$ divides $f(x)$ and we write $g(x) \mid f(x)$. Otherwise, $g(x) \nmid f(x)$. In this case, we say that $g(x)$ is a factor of $f(x)$. If a polynomial has no nonconstant polynomial factor of smaller degree, we say that the polynomial is irreducible.
%Double check this definition of irreducible.
%Should we talk about $0 \mid 0$? Should we redefine what it means to divide?
\ic{Note here that we're generalizing the definition of $\mid$. This reduces to the definition we had for integers.}
\end{enumerate}
\ex Show over $\mathbb{R}$ that
\[
(x-1) \nmid (x^{2} +1)
\]
\pf By DAP, there exists $q(x)$ and $r(x)$ polynomials over $\mathbb{R}$ such that
\[
x^{2} + 1 = (x-1)q(x) + r(x)
\]
To show that $r(x) \neq 0$, it suffices to show that $r(a) \neq 0$ for some $a \in \mathbb{F}$. Take $x=1$. Then 
\[
(1)^{2} + 1 = (1-1)q(1) + r(1)
\]
giving $2=r(1)$. Therefore, $r(x) \neq 0$ hence $(x-1)\nmid x^{2}+1$. \qed
\ic{My guess is that you will need to push this to the next lecture which is fine.}
\textbf{Long Division}
\vskip1em
Let's divide
\begin{align*}
f(z) = iz^{3} + (i+3)z^{2} + (5i+3)z + (2i-2)
\end{align*}
by $g(z) = z + (i+1)$.
\begin{center}
\includegraphics[scale=0.5]{LongDiv1.png}
\end{center}


