# Friday January 10

Recall that $\CC$ is a field, where $$z = x + iy \implies \bar z = x - iy$$ and if $z\neq 0$ then $$z\inv = \frac{\bar z} {\abs{z}^2}$$

Lemma (Triangle Inequality)
: 	$$\abs{z + w} \leq \abs z + \abs w.$$

Proof
: 
	\begin{align*}
	(\abs z + \abs w)^2 - \abs{z+w}^2 = 2( \abs{z\bar w} - \Re z\bar w ) \geq 0
	.\end{align*}

Lemma (Reverse Triangle Inequality)
: $$\abs{\abs z - \abs w} \leq \abs{z-w}.$$

Proof
: 
	\begin{align*}
	\abs z = \abs{z-w + w} \leq \abs{z-w} + \abs w \implies \abs w - \abs z \leq \abs{z-w} = \abs{w-z}
	.\end{align*}

Fact
: $(\CC, \abs{\wait})$ is a normed space.

Definition (Limits of Complex Sequences)
: $$\lim z_n = z \iff \abs{z_n - z} \to 0 \in \RR.$$

Definition (Complex Discs)
: 	A *disc* is defined as $D_r(z_0) \definedas \theset{z\in\CC \suchthat \abs{z-z_0} < r}$, and a subset is open iff it contains a disc.
	By convention, $D_r$ denotes a disc about $z_0 = 0$.

Definition (Convergence in $\CC$)
: $\sum_k z_k$ *converges* iff $S_N \definedas \sum_{\abs k < N} z_k$ converges.

Note that $z_n \to z$ and $z_n = x_n + iy_n$, and $$\abs{z_n - z} = \sqrt{(x_n - x)^2 - (y_n - y)^2} < \varepsilon \implies \abs{x - x_n}, \abs{y - y_n} < \varepsilon.$$

Since $\RR$ is complete iff every Cauchy sequence converges iff every bounded monotone sequence has a limit.

> Note: This is useful precisely when you don't know the limiting term.

Note that $\sum_k z_k$ thus converges if $\abs{\sum_{k=m}^n z_k} < \varepsilon$ for $m, n$ large enough, so sums converges iff they have small tails.

Definition (Absolute Convergence)
: 	$S_N = \sum^N z_k$ *converges absolutely* iff $\tilde S \definedas \sum^N \abs{z_k}$ converges.

Note that the partial sums $\sum^N \abs{z_k}$ are monotone, so $\tilde S_N$ converges iff the partial sums are bounded above.

Definition (Power Series)
: 	A sum of the form $\sum_{k=0}^\infty a_k z_k$ is a *power series*.

*Examples*:

\begin{align*}
\sum x^k &= \frac 1 {1-x} \\
\sum (-x^2)^k &= \frac 1 {1+x^2}
.\end{align*}

Note that both of these have a radius of convergence equal to 1, since the first has a pole at $x=1$ and the second as a pole at $x = i$.