\newlec

\ic{There's a large probability that you might have extra time in this lecture - there are ways to fill that time in later lectures with some extra complex numbers proofs.}

%You don't need to discuss rings and fields here. We did that earlier.

\textbf{Complex Numbers}

Our current view of important sets:
\[
\mathbb{N} \subseteq \mathbb{Z} \subseteq \mathbb{Q} \subseteq \mathbb{R}
\]
These sets can be thought of as helping us to solve polynomial equations. However, $x^{2} +1 =0$ has no solution in any of these sets.
\ic{This is the 3 minute mark}
\defn A complex numbers (in standard form) is an expression of the form $x + yi$ where $x,y \in \mathbb{R}$ and $i$ is the imaginary unit. Denote the set of complex numbers by 
\[
\mathbb{C} := \{x + yi : x,y \in \mathbb{R}\}
\]
\ex $1 + 2i$, $3i$, $\sqrt{13} + \pi i$, $2$ (or $2 + 0i$).
\vskip1em
\note \begin{enumerate}
\item $\mathbb{R} \subseteq \mathbb{C}$
\item If $z = x + yi$, then $x = \text{Re}(z) = \Re(z)$ is called the real part and $y = \text{Im}(z) = \Im(z)$ is called the imaginary part.
\end{enumerate}
\defn Two complex numbers $z = x + yi$ and $w = u + vi$ are equal if and only if $x = u$ and $y = v$.
\vskip1em
\defn A complex number $z=x+yi$ is...
\begin{enumerate}
\item Purely real (or simply real) if $\Im(z) = 0$, that is, $z=x$
\item Purely Imaginary if $\Re(z) = 0$, that is, $x = yi$.
\end{enumerate}
We turn $\mathbb{C}$ into a commutative ring by defining operations as follows:
\begin{enumerate}
\item $(x +yi) \pm (u + vi) := (x \pm u) + (y \pm v)i$
\item $(x + yi)(u + vi) := (xu - vy) + (xv + uy)i$
\end{enumerate}
By this definition, we have
\[
i^{2} = i \cdot i = (0 + i)(0 + i) = -1 + 0i = -1.
\]
Therefore, $i$ is a solution of $x^{2} + 1$. With this in mind, you can remember multiplication just by multiplying terms as you would with polynomials before.
\begin{align*}
(x + yi)(u + vi) = xu + xvi + yiu + yivi = xu + (xv + yu)i + yvi^{2} = xu-yv + (xv + uy)i
\end{align*}
\ex 
\begin{enumerate}
\item $(1 + 2i) + (3 + 4i) = 4 + 6i$
\item $ (1 + 2i) - (3 + 4i) = -2-2i$
\item $(1 + 2i)(3 + 4i) = 3-8 + (4+6)i = -5 + 10i$
\end{enumerate}
We note that $\mathbb{C}$ is a field by observing that the multiplicative inverse of a nonzero complex numbers is
\[
(x + yi)^{-1} = \frac{x}{x^{2} + y^{2}} - \frac{y}{x^{2} + y^{2}}i
\]
\exr If $z \in \mathbb{C}$ and $z \neq 0$, then $z \cdot z^{-1} = 1$
\ic{This is the 20-25 minute mark.}
For complex numbers $u,v,w,z$ with $v$ and $z$ nonzero, the above is consistent with the usual fraction rules:
\[
\frac{u}{v} + \frac{w}{z} = \frac{uz + vw}{vz} \quad\quad \text{and} \quad\quad \frac{u}{v} \cdot \frac{w}{z} = \frac{uw}{vz}
\]
For $k \in \mathbb{N}$ and $z \in \mathbb{C}$, define
\[
z^{0} = 1 \quad \quad z^{1}=z \quad\quad z^{k+1} = z\cdot z^{k}
\]
and further that $z^{-k} := (z^{-1})^{k}$. With these definitions, the usual exponent rules hold, namely
\[
z^{m+n} = z^{m} \cdot z^{n} \quad\quad (z^{m})^{n} = z^{mn}
\]
for $m,n\in\mathbb{Z}$.
\vskip1em
\ex Write $\frac{1 + 2i}{3 - 4i}$ in standard form.
\vskip1em
\soln 
\begin{align*}
\frac{1 + 2i}{3 - 4i} &= (1 + 2i)(3 - 4i)^{-1}\\
&= (1 + 2i)\left(\frac{3}{3^{2} + 4^{2}} - \frac{(-4)}{3^{2} + 4^{2}}i \right)\\
&= (1 + 2i) \left( \frac{3}{25} + \frac{4}{25}i\right)\\
&= \frac{3}{25} - \frac{8}{25} + \left( \frac{4}{25} + \frac{6}{25}\right)i\\
&= \frac{-5}{25} + \frac{10}{25}i\\
&= \frac{-1}{5} + \frac{2}{5}i
\end{align*}
\ic{This is the 30 minute mark}
\newpage
\hdc

Express the following in standard form
\begin{enumerate}
\item $z = \frac{(1 - 2i) - (3 + 4i)}{5 - 6i}$
\item $w = i^{2015}$
\end{enumerate}

\vskip3em

\soln
\begin{enumerate}
\item 
\begin{align*}
z &= ((1 - 2i) - (3 + 4i))(5-6i)^{-1}\\
&= (-2 - 6i)\left(\frac{5}{5^{2}  +6^{2}}-\frac{(-6)}{5^{2} + 6^{2}}i\right)\\
&= (-2 - 6i)\left(\frac{5}{61}+\frac{6}{61}i\right)\\
&= \frac{-10}{61} + \frac{36}{61} + \left( \frac{-12}{61} - \frac{30}{61}\right)i\\
&= \frac{26}{61} - \frac{42}{61}i
\end{align*}
\item Recall that $i^{2} = -1$ and $i^{4} = 1$. Thus,
\begin{align*}
w &= i^{2015} \\
&= (i^{4})^{503} \cdot i^{3}\\
&= 1^{503} \cdot i^{2} \cdot i\\
&= -i
\end{align*}
\end{enumerate}

\ic{This is the 40 minute mark - you can easily go on to the next lecture or use this time to catch up.}
