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+\def\Re{\mathop{\rm Re}\nolimits}
+\def\Im{\mathop{\rm Im}\nolimits}
+\def\Log{\mathop{\rm Log}\nolimits}
+\let\rule\hrule
+\def\hrule{\medskip\rule\medskip}
+
+%page 170
+% remember the cauchy integral proof
+\noindent{\bf 4.}
+
+Inside the contour, the extension of the Cauchy integral applies and
+tells us that $g(z)$ is equivalent to $2\pi if''(z)/2!$ where
+$f(z) = z^3 + 2z.$ $f''(z) = 6z,$ so this evaluates to $6\pi iz.$
+
+The integrand is analytic where $s\neq z,$ so with $z$ outside of $C,$
+the integral is zero by Cauchy's integral theorem---since the integrand
+is analytic inside and on the simply connected region enclosed by $C.$
+
+\noindent{\bf 6.}
+
+To determine whether $g(z)$ is analytic, we can use the definition of
+the derivative
+$$g'(z) = {g(z+\Delta z) - g(z)\over \Delta z}.$$
+If it exists, $g$ is analytic.
+$$g'(z) = {1\over 2\pi i}\int_C {f(s)\over s-z-\Delta z} - {f(s)\over
+s-z} {ds\over \Delta z}$$$$
+= {1\over 2\pi i}\int_C {f(s)ds\over (s-z-\Delta z)(s-z)}
+= {1\over 2\pi i}\int_C {f(s)ds\over (s-z)^2} + {1\over 2\pi i}\int_C
+{\Delta zf(s) ds\over (s-z-\Delta z)(s-z)^2},$$
+and that second term is limited by zero because $\Delta z$ can be
+arbitrarily low.
+
+If $f(z)$ is continuous on $C,$ where $C\cap\{z\} = \emptyset,$
+$|s-z|>0,$ so ${f(z)\over (s-z)^2}$ is continuous and the derivative
+exists.
+
+%page 177
+\noindent{\bf 2.}
+
+Because $f(z) \neq 0,$ and $f$ is analytic, $g(z) = 1/f(z)$ is
+continuous and also analytic.
+By the result for maximum values, there exists a value $M$ such that
+$|g(z)| \leq M$ for all $z\in R.$
+Because $f(z)$ is not constant, $g(z)$ also isn't constant, and the only
+points such that $g(z_0) = M$ are on the boundary.
+
+These conditions imply that, for all $z\in R,$ $|f(z)| \geq 1/M,$ and
+the only points such that $f(z_0) = 1/M$ is on the boundary.
+This proves that the minimum of $f$ are on the boundary.
+
+\noindent{\bf 4.}
+
+$$|f(z)|^2 = |\sin z|^2 = \sin^2 x + \sinh^2 y$$
+is maximized when $\sin x$ is maximized and when $\sinh y$ is maximized.
+$\sin x$ has a maximum at $x=\pi/2$ on $(0,\pi)$ because this is the
+maximum of the real function.
+$\sinh y$ has a maximum at $y=1$ on $(0,1)$ because $\sinh x$ tends to
+infinity as $x$ approaches either positive or negative infinity.
+Therefore, $|f(z)|$ is maximized at $x + yi = \pi/2 + i.$
+
+%page 185
+\noindent{\bf 2.}
+
+The principal arguments of these numbers are, for $\Theta_{2n},$
+$\tan^{-1}(1/n^2),$ and for $\Theta_{2n+1},$ $\tan^{-1}(-1/n^2).$
+Both of these are arbitrarily close to 0 for sufficiently large $n,$ so
+the series has sum $0.$
+
+This series converges unlike Example 2 in Section 60 because the value
+to which it converges is not a branch cut, where a given point can be
+infinitely close to distinct values of $\Theta.$
+
+\noindent{\bf 3.}
+
+If $$\lim_{n\to\infty} z_n = z,$$
+
+$|z_{n} - z| < \epsilon$ for $n > n_0.$
+Because $||z_n| - |z|| \leq |z_n - z|,$ $||z_n| - |z|| < \epsilon$ also
+for $n > n_0.$
+
+This shows that
+$$\lim_{n\to\infty} |z_n| = |z|.$$
+
+\noindent{\bf 8.}
+
+$$\sum_{n=1}^\infty z_n = S = S_x + iS_y,$$
+which are in turn defined as infinite sums over the real numbers.
+Similarly, $T = T_x + iT_y.$
+
+Therefore,
+$$\sum_{n=1}^\infty (z_n + w_n) = \sum_{n=1}^\infty (z_{x_n} + iz_{y_n}
++ w_{x_n} + iw_{y_n}) = \sum_{n=1}^\infty (z_{x_n} + w_{x_n}) +
+i(z_{y_n} + w_{y_n})$$$$ = (S_x + T_x) + i(S_y + T_y) = S_x + iS_y + T_x
++ iT_y = S + T,$$
+made possible by the analogous property for real series and the fact
+that $\sum_{n=1}^\infty iz_n = i\sum_{n=1}^\infty z_n.$
+
+\bye