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\centerline{\bf Class \#7 (Wednesday January 28) }
\begin{description}
\item[SUMMARY] Limits of Complex Functions and the Point At Infinity
\vspace{-12pt}
\item[CURRENT READING] Brown \& Curchill, pages 33-40
\vspace{-12pt}
\item[NEXT READING] Brown \& Curchill, pages 40-45
\end{description}
 
\underline{{\bf Update} on {\it Class \#5}}\\
Yes, closed sets can be connected. Open connected sets are more interesting because they are also called {\bf domains} or open regions. If a set is closed and connected it's called a closed region.\\ 

If a set does not have any limit points, such as the set consisting of the point $\lbrace 0 \rbrace$, then it is {\bf closed}. [It contains all its limit points (it just doesn't have any limit points).]\\

Remember, if a set contains all its boundary points (marked by solid line), it is {\bf closed.} \\
If a set contains none of its boundary points (marked by dashed line), it is {\bf open.}\\

Also, {\it sets can be both open and closed.} An example is the set ${\cal C}$ (the Complex Plane). It has no boundary points. Thus $\C$ is closed since it contains all of its boundary points (doesn't have any) and $\C$ is open since it doesn't contain any of its boundary points (doesn't have any).\\

Also, {\it sets can be neither open or closed.} The set $0<|z| \le 1$ has two boundaries (the set $|z|=1$ and the point $z=0$). It contains the first boundary ($|z|=1$), so it is not open, but it does not contain the boundary point $z=0$ so it is not closed.
$z = 0$ is also a limit point for this set which is not in the set, so this is another reason the set is not closed.\\

{\large\bf Limits}\\
Suppose that $f(z)$ is defined on a deleted neighborhood of $z_0$. In order to say that ${\displaystyle \lim_{z \rightarrow z_0} f(z) = w_0}$ we must be able to show that

$$\forall \epsilon > 0, \quad \exists \delta > 0 \ni |z - z_0| < \delta \Rightarrow |f(z) - w_0| < \epsilon$$

This may look like dense mathematical language, but in english this means that for every positive number $\epsilon$ (no matter how small) there exists a number $\delta$ (which depends on the choice of $\epsilon$) so that regardless of how close you get to the point $z_0$ in the deleted neighborhood around it in the $z$-plane  you can also get arbitrarily close to the value $w_0$ in the $w$-plane.\\

\newpage
{\bf Rules on Limits}\\
The rules on limits of complex functions are identical to the rules for limits of real functions of real variables (as you'd expect)

Suppose that
$\lim_{z \rightarrow z_{\scriptscriptstyle 0}} f(z) = w_0 \mbox{ and } \lim_{z \rightarrow z_{\scriptscriptstyle 0} } F(z) = W_0$
then
\begin{eqnarray*}
\lim_{z \rightarrow {z_0}} [ f(z) + F(z)] & =  &w_0 + W_0 \\
\lim_{z \rightarrow \z0} [ f(z) F(z)] & = & w_0  W_0\\
\lim_{z \rightarrow \z0}  {\displaystyle \frac{f(z)}{F(z)}} & = & { w_0 \over  W_0 } \qquad (W_0 \neq 0) \\
\lim_{z \rightarrow \z0} |f(z)| & = & |w_0| \\
\lim_{z \rightarrow \z0} c  & = & c\\
\lim_{z \rightarrow \z0} z^n & = & \z0^n\\
\lim_{z \rightarrow \z0} P(z) & = & P(\z0) \qquad ( \mbox{where } P(z) \mbox{ is a polynomial})
\end{eqnarray*}

IF
$f(z) = u(x,y) + i v(x,y), \qquad \z0 = x_0 + i y_0 \qquad \mbox { and } w_0 = u_0 + i v_0$
THEN

${\displaystyle \lim_{z \rightarrow \z0}} f(z) = w_0$ if and only if
${\displaystyle \lim_{(x,y) \rightarrow (\x0,\y0)}} u(x,y) = u_0 
\mbox{ and } 
{\displaystyle \lim_{(x,y) \rightarrow (\x0,\y0)}} v(x,y) = v_0$\\

\underline{\bf Examples}
\begin{itemize}
\item[(a)] ${\displaystyle \lim_{z \rightarrow {1+2i}} }2 |z| + i z^2 + 2.5 - .1 i =$
\item[(b)] ${\displaystyle \lim_{z \rightarrow 3 \pi i}} z e^{z} = $
\item[(c)] ${\displaystyle \lim_{z \rightarrow 0} } {\displaystyle \frac{z^8+z^4+z^2+z-1}{z^3+4z^3-9}} =$
\item[(d)] ${\displaystyle \lim_{z \rightarrow 1-i}} 2 x y - i x^2 -i y^2 =$
\end{itemize}

{\bf\large Point at Infinity}\\
When dealing with real numbers we often speak of two different concepts, denoted $\boldmath -\infty \unboldmath$ and $\boldmath +\infty \unboldmath$.

However, the complex number infinity is represented as one particular point in the Argand plane. We call this the {\bf point of infinity} and we rename the Argand plane the {\bf extended $z$ plane} or the {\bf extended complex plane} when we include it. 

The point at infinity can be considered to be the image of the origin $z=0$ under the mapping $w=1/z$

To compute complex limits involving $\infty$ we use this idea:

$${\displaystyle \lim_{z \rightarrow \infty} f(z) = \lim_{w \rightarrow 0} f \left( \frac{1}{w} \right) }$$

In order for us to say a complex function $f(z)$ becomes unbounded at a point $z_0$, (i.e. $f(z_0) = \infty$) we must have that
$$\forall M > 0, \quad  \exists \delta > 0 \quad  \ni 0 < |z-z_0| < \delta \Rightarrow |f(z)| > M$$

\underline{ Examples}\\
\begin{itemize}
\item[(a)] ${\displaystyle \lim_{z \rightarrow \infty} \frac{2 z + 3}{3 z - i} } = $
\item[(b)] ${\displaystyle \lim_{ z \rightarrow \infty}  z^2 + 3 z - 5 i + 4} = $
\end{itemize}
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