\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