\def\Res{\mathop{\rm Res}}
\let\rule\hrule
\def\hrule{\medskip\rule\medskip}

%page 237
\noindent{\bf 3.}

With
$$f(z) = {4z-5\over z(z-1)},$$

$$\int_C f(z)dz = -2\pi i\Res_{z=\infty}f(z) =
2\pi i\Res_{z=0}{1\over z^2}f\left({1\over z}\right) =
2\pi i\Res_{z=0}{z(4 - 5z) \over z^2(1 - z)}.$$

$${z(4-5z)\over z^2(1-z)} = (4/z-5)(1+z+z^2+\cdots) \to b_1 = 4.$$
Therefore, the value of the integral is $8\pi i.$

\hrule
%page 242
\noindent{\bf 1.}

\noindent{\it (a)}

$$f(z) = z\exp\left({1\over z}\right) = z\left(1 + {1\over z} +
{1\over 2z^2} + \cdots\right),$$
giving principal part
$${1\over 2z} + {1\over 6z^2} + \cdots,$$
meaning that $z=0$ is an essential singular point.

\noindent{\it (b)}

$$f(z) = {z^2\over 1+z} = {z^2\over 1 - (-z)} = z^2 - z + 1 - {1\over z}
+ {1\over z^2} + \cdots,$$
giving principal part
$$-{1\over z} + {1\over z^2} + \cdots,$$
demonstrating an essential singular point at $z=0.$

\noindent{\it (c)}

$$f(z) = {\sin z\over z} = 1 - {z^2\over 3!} + {z^4\over 5!} - {z^6\over
7!}.$$
The principal part is $0,$ so $z=0$ is a removable singular point.

\noindent{\it (d)}

$$f(z) = {\cos z\over z} = {1\over z} - {z\over 2!} + {z^3\over 4!} +
\cdots,$$
giving principal part $1/z,$ meaning $z=0$ is a simple pole.

\noindent{\it (e)}

$$f(z) = {1\over (2-z)^3} = -{1\over (z-2)^3}$$
is a complete Laurent series, and also its principal part, telling us
there is a pole of order 3 at $z=2.$

\noindent{\bf 3.}

\noindent{\it (a)}

If $f(z_0) \neq 0,$ $f(z)$ has Taylor series representation $a_0 +
{a_1(z-z_0)\over 1!} + {a_2(z-z_0)^2\over 2!} + {a_3(z-z_0)^3\over 3!}
+ \cdots.$ Therefore, $g(z)$ has Laurent series representation
${a_0\over z-z_0} + a_1 + {a_2(z-z_0)\over 2!} + {a_3(z-z_0)^3\over 3!},$
evidencing a simple pole.

\noindent{\it (b)}

Similarly, if $f(z_0) = 0,$ $g(z)$ has Taylor series ${a_1\over 1!} +
{a_2(z-z_0)\over 2!} + {a_3(z-z_0)^2\over 3!},$ with principal part $0$
and a removable singular point.

\hrule
%page 246
\noindent{\bf 3.}

\noindent{\it (a)}

$\sinh z/z = f(z)$ is non-zero and analytic at $z=0,$ meaning that
$g(z) = \sinh z/z^4 = f(z)/z^3$ has a pole of order 3 at $z=0.$
The Laurent series of $\sinh z/z^4$ is ${1\over z^3} + {1\over 3!z} +
{z\over 5!} + \cdots,$ giving a residue of $1/6.$

\noindent{\it (b)}

$${1\over z(e^z - 1)} = {1\over z(1+z+{z^2\over 2}+{z^3\over
6}+\cdots-1)} = {1\over z^2(1 + {z\over 2} + {z^2\over 6} + \cdots)} =
{1\over z^2}{1\over 1 - (-{z\over 2} - {z^2\over 6} - \cdots)},$$
and by the standard $1/(1-z)$ expansion since this series is less than 1
(it equals ${e^z - 1\over z}-1,$ which is near 0 about $z=0,$ which can
be verified by L'hospital's rule),
$$ = {1\over z^2} - {1\over 2z} - {1\over 6} + \cdots$$

This is a residue of $-1/2.$

\noindent{\bf 5.}

\noindent{\it (a)}

$|z|=2$ contains the singular point at $z=0,$ and $f(z)$ can be defined
as
$$\phi(z)\over z^3$$ where $\phi(z) = 1/(z+4),$ which is analytic within
the region. The residue is $\phi''(0)/2! = {2/(0+4)^3\over 2} = 1/64.$
This gives the integral a value of $2\pi i/64 = \pi i/32.$

\noindent{\it (b)}

This region contains both singularities at $z=0$ and $z=-4,$ and its
residue is the sum of theirs.
We have already evaluated the residue at $z=0,$ so we will now evaluate
the residue at $z=-4.$
$$f(z) = {{1\over z^3}\over z+4} = {\phi(z)\over z+4}.$$
This has residue $\phi(-4) = -1/64.$
With the residue at $z=0,$ this gives a sum total of $0.$

\noindent{\bf 6.}

$C$ contains singularities at $-i,$ $0,$ and $i.$
At $z=0,$
$$f(z) = {\phi(z)\over z},$$
with
$$\phi(z) = {\cosh \pi z\over z^2+1}.$$
This tells us the residue is $\phi(0) = \cosh 0/1 = 1.$

Similarly, $z=i$ has
$$f(z) = {\phi(z)\over z-i},$$
with
$$\phi(z) = {\cosh \pi z\over z(z+i)} \to \phi(i) =
{\cosh(\pi i)\over -2} = {1\over 2}.$$

And $z=-i$ has
$$f(z) = {\phi(z)\over z+i},$$
with
$$\phi(z) = {\cosh \pi z\over z(z-i)} \to \phi(i) =
{\cosh(-\pi i)\over 2} = {1\over 2}.$$

The sum residue is $2,$ giving the integral a value of $2\cdot 2\pi i =
4\pi i.$

\iffalse
$C$ contains all singularities, so the integral can be determined as
$2\pi i\Res_{z=\infty} f(z).$

$${1\over z^2}f\left({1\over z}\right) =
{z^3\cosh(\pi/z) \over 1+z^2} =
{z^3\over 1+z^2}\left(1 + {(\pi/z)^2\over 2!} + {(\pi/z)^4\over
4!}\right) =

%{1\over 1+z^2}\left(z^3 + {z\pi^2\over 2!} + {\pi^4\over 4!z} +
%{\pi^6\over 6!z^3} + \cdots\right)
$$
\fi

\bye
