Some minor fixes in language, fix figure to be in line with the book

ca.tex

@@ -15106,13 +15106,13 @@ \subsection{Schwarz reflection principle}
 =
 0 .
 \end{equation*}
-In other words, the mean value equals $F(z) = 0$,
+That is, the mean value equals $F(z) = 0$,
 and the mean-value property is satisfied for all small enough
 $r$ at every $z \in L$.
 We proved above that $F$ is harmonic on $U \setminus L$ and
 so the mean-value property
-is satisfied for all small enough
-$r$ around any of $U \setminus L$ as well.
+is also satisfied for all small enough circles
+around any $z \in U \setminus L$.
 By the mean-value property (\thmref{thm:meanprop}), $F$ is harmonic in $U$.
 \end{proof}
 
@@ -15811,9 +15811,9 @@ \section{Infinite products}
 \label{sec:prod}
 
 If a
-function has zeros at $0$, $1$, and $i$ we
+function has zeros at $0$, $1$, and $i$, we
 can write $f$ as
-$f(z) = g(z) z(z-1)(z-i)$.  And if those are the only zeros, then 
+$f(z) = g(z) z(z-1)(z-i)$.  If those are the only zeros, then 
 $g$ is never zero.  If there are infinitely many zeros, however,
 things become difficult.
 Can we factor out the zeros out of $\sin z$?  Can we
@@ -15825,7 +15825,7 @@ \section{Infinite products}
 we take care of convergence.\footnote{Beware of
 formal expressions bearing gifts.
 Especially ones with infinite sets in them.
-For example, $\prod_{n \in \Z} (z-\pi n)$ does not make sense.}
+For example, $\prod_{n \in \Z} (z-\pi n)$ does not actually make any sense.}
 Of course, we need to first
 figure out what we mean by convergence.  Once we figure that out,
 we will show that convergence happens the way we want to
@@ -15900,7 +15900,7 @@ \section{Infinite products}
 
 For the other direction, suppose that the sequence of partial products
 converges.  For all but finitely many $n$, $\sabs{a_n} < 1$.
-Otherwise the product would double infinitely often and go to
+Otherwise the partial products would double infinitely often and go to
 infinity.  So suppose that $\sabs{a_n} < 1$ for all $n \geq N$.
 If $\sabs{a_n} < 1$, then $\sabs{a_n} \leq 2 \log \bigl(1+\sabs{a_n}\bigr)$,
 and
@@ -15917,7 +15917,7 @@ \section{Infinite products}
 
 An immediate consequence is that if 
 $\prod_{n=1}^\infty (1+a_n)$ converges absolutely, then
-$\{ a_n \}$ converges to $0$.  Actually much more than that, they have to
+$\{ a_n \}$ converges to $0$.  Not only that, they
 go to zero fast enough to be absolutely summable.
 
 As for series, we need to know that absolute convergence really is convergence.
@@ -15959,11 +15959,11 @@ \section{Infinite products}
 \\
 & \leq
 \sum_{n=k+1}^m 
-\left(
+\Bigl(
 \log \sabs{1+a_n}
 +
 \abs{ \Arg (1+a_n)}
-\right) .
+\Bigr) .
 \end{split}
 \end{equation}
 As before,
@@ -15975,7 +15975,8 @@ \section{Infinite products}
 \end{equation*}
 and the series on the right converges via \propref{prop:infprodabsconvinfsum}.
 Next, $\Arg(1+a_n)$ is between $\nicefrac{-\pi}{2}$ and
-$\nicefrac{\pi}{2}$, and $a_n$ is going to zero.  So for all $n$ large
+$\nicefrac{\pi}{2}$, and $a_n$ goes to zero.  A calculus exercise
+shows that for all $n$ large
 enough, $\babs{\Arg(1+a_n)} \leq 2\sabs{a_n}$.  As $\sum
 \sabs{a_n}$ converges, 
 $\sum \babs{\Arg (1+a_n)}$ converges.
@@ -16109,9 +16110,8 @@ \section{Infinite products}
 
 \begin{exbox}
 \begin{exercise}
-Suppose $\sabs{1+a_n} \leq r < 1$ for all
-$n$, 
-prove that $\prod_{n=1}^\infty (1+a_n)$ converges to $0$, but that the
+Suppose $\sabs{1+a_n} \leq r < 1$ for all $n$.
+Prove that $\prod_{n=1}^\infty (1+a_n)$ converges to $0$, but that the
 convergence is not absolute.
 \end{exercise}
 
@@ -17013,7 +17013,7 @@ \section{Runge's theorem}
 \end{exercise}
 \end{exbox}
 
-Let us now approximate simple poles of the form $\frac{1}{z-p}$ by rational
+We now approximate simple poles of the form $\frac{1}{z-p}$ by rational
 functions with poles in a given set.  This procedure is called
 \emph{\myindex{pole pushing}} as we are going to \myquote{push} the poles along a
 path to where we need them to be.

changes-draft.html

@@ -83,4 +83,5 @@
 <li><b>Simplify Exercise 7.2.26 a little</b> by assuming that \(U\) is
   connected to having to think about the technicality of countably many
   components which is not really important.
+<li>In Figure 7.6 stop marking \(U_-,\) which we never defined.
 <li>Clarify the proof of Rado's theorem.

figures/schwarzreflection.fig

@@ -1,4 +1,4 @@
-#FIG 3.2  Produced by xfig version 3.2.7a
+#FIG 3.2  Produced by xfig version 3.2.8b
 Landscape
 Center
 Inches
@@ -9,8 +9,8 @@ Single
 1200 2
 0 32 #b9b9b9
 0 33 #e6e6e6
-1 3 0 1 0 0 50 -1 20 0.000 1 0.0000 900 375 18 18 900 375 918 375
 1 3 0 1 0 0 50 -1 20 0.000 1 0.0000 900 975 18 18 900 975 918 975
+1 3 0 1 0 0 50 -1 20 0.000 1 0.0000 900 375 18 18 900 375 918 375
 2 1 0 1 0 7 50 -1 -1 0.000 0 0 -1 0 0 2
 	 225 675 2925 675
 2 1 0 3 0 7 50 -1 -1 0.000 0 0 -1 0 0 2
@@ -23,7 +23,6 @@ Single
 	 0.000 -0.500 -0.500 -0.500 -0.500 0.000
 4 1 0 50 -1 0 12 0.0000 2 150 330 2250 876 $L$\001
 4 1 0 50 -1 0 12 0.0000 2 150 405 2775 876 $\\R$\001
-4 1 0 50 -1 0 12 0.0000 2 150 540 1500 300 $U^+$\001
-4 1 0 50 -1 0 12 0.0000 2 150 495 1500 1200 $U^-$\001
+4 1 0 50 -1 0 12 0.0000 2 165 555 1500 300 $U_+$\001
 4 0 0 50 -1 0 12 0.0000 2 150 300 975 450 $z$\001
 4 0 0 50 -1 0 12 0.0000 2 165 795 975 1050 $\\bar{z}$\001

figures/schwarzreflection.pdf_t

@@ -14,9 +14,7 @@
 }}}}
 \put(2776,-37){\makebox(0,0)[b]{\smash{{\SetFigFont{12}{14.4}{\familydefault}{\mddefault}{\updefault}{\color[rgb]{0,0,0}$\R$}%
 }}}}
-\put(1501,539){\makebox(0,0)[b]{\smash{{\SetFigFont{12}{14.4}{\familydefault}{\mddefault}{\updefault}{\color[rgb]{0,0,0}$U^+$}%
-}}}}
-\put(1501,-361){\makebox(0,0)[b]{\smash{{\SetFigFont{12}{14.4}{\familydefault}{\mddefault}{\updefault}{\color[rgb]{0,0,0}$U^-$}%
+\put(1501,539){\makebox(0,0)[b]{\smash{{\SetFigFont{12}{14.4}{\familydefault}{\mddefault}{\updefault}{\color[rgb]{0,0,0}$U_+$}%
 }}}}
 \put(976,389){\makebox(0,0)[lb]{\smash{{\SetFigFont{12}{14.4}{\familydefault}{\mddefault}{\updefault}{\color[rgb]{0,0,0}$z$}%
 }}}}