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\usepackage{epsfig}
\begin{document}
\newcommand\ds{\displaystyle}
\parindent=0pt

QUESTION

Expand the following functions in Taylor or Laurent
series about each of the points $z=0$ and $z=i$
\begin{description}

\item[(a)]
$\frac{1}{1+z}$

\item[(b)]
$\frac{1}{z^2(z-i)}$

\item[(c)]
$\frac{\cosh z}{1+z^2}$

\end{description}

\bigskip

ANSWER
\begin{description}


  \item[(a)]
  \begin{description}
  \item[(i)]
  $$ \ $$
  $\begin{array}{l}
  \ds\frac{1}{1+z}=\frac{1}{1-(-z)}=1+(-z)+(-z)^2+\ldots\\
  =1-z+z^2-z^3+\ldots\\
  \textrm{for  }|z|<1.
  \end{array}
  \ \ \
  \begin{array}{c}
  \epsfig{file=272-1-4.eps, width=25mm}
  \end{array}$

  \item[(ii)]
  $$ \ $$
  $\begin{array}{l}
  \ds\frac{1}{1+z}=\frac{1}{1+i+(z-1)}\\
  \ds\frac{1}{1+i}\frac{1}{1+\frac{z-i}{1+i}}\\
  \ds\frac{1}{1+i}(1-\frac{z-i}{1+i}+(\frac{z-i}{1+i})^2-(\frac{z-i}{1+i})^3+\ldots
  )\\
  \ds\textrm{for }|\frac{z-i}{1+i}| <1 \Longleftrightarrow |z-i| \leq
  \sqrt{2}
  \end{array}
  \ \ \
  \begin{array}{c}
  \epsfig{file=272-1-5.eps, width=30mm}
  \end{array}$

  \end{description}

  \item[(b)]

  \begin{description}

  \item[(i)]
  \begin{eqnarray*}
  \frac{1}{z^2(z-i)}&=&\frac{1}{z^2}\frac{1}{-i}\frac{1}{1+\frac{z}{-i}}\\
  &=&\frac{i}{z^2}\frac{1}{1+iz}\\
  &=&\frac{i}{z^2}[1-iz+(iz)^2-(iz)^3+(iz)^4-\ldots]\\
  &=&\frac{i}{z^2}[1-iz-z^2+iz^3+z^4-\ldots]\\
  &=&iz^{-2}+z^{-1}-i-z+iz^2+\ldots\\
  & &\textrm{ for } 0 <|z|<1
  \end{eqnarray*}

  $\begin{array}{l}
  \textrm{The principal part is }iz^{-2}+z^{-1}\\
  \textrm{Res}=1.\\
  \end{array}
  \ \ \
  \begin{array}{c}
  \epsfig{file=272-1-6.eps, width=30mm}
  \end{array}$

  \item[(ii)]
  \begin{eqnarray*}
   \frac{1}{z^2(z-i)}&=&\frac{1}{z-i}\left[\frac{1}{(z-i)+i}\right]^2\\
   &=&\frac{1}{z-i}\left[\frac{1}{i}\frac{1}{1+\frac{z-i}{i}}\right]^2
   \textrm{ for } 0<|z-i|<1\\
   &=&-\frac{1}{z-i}\left[1-\frac{z-i}{i}+\left(\frac{z-i}{i}\right)^2
   -\left(\frac{z-i}{i}\right)^3+\ldots \right]^2\\
   &=&-\frac{1}{z-i}\left[1-2\frac{z-i}{i}+\ldots\right]\\
   &=&-(z-i)^{-1}-2i+\ldots
  \end{eqnarray*}
  Res=$-1$

  \begin{center}
  \epsfig{file=272-1-7.eps, width=30mm}
  \end{center}


  \end{description}

  \item[(c)]

  \begin{description}

  \item[(i)]
  $$ \ $$
  $\begin{array}{l}
  \ds\frac{\cosh z}{1+z^2}\\
  =(1+\frac{z^2}{2!}+\frac{z^4}{4!}+\ldots)(1-z^2+z^4-\ldots)
  \end{array}
  \ \ \
  \begin{array}{c}
  \epsfig{file=272-1-8.eps, width=25mm}
  \end{array}$

  This is the Taylor series of cosh (valid for all $x$) times the
  geometric series in $(-z^2)$ (valid for $|z^2|<1
  \Longleftrightarrow|z|<1$.) Hence
  $$\frac{\cosh
  z}{1+z^2}=1+(\frac{1}{2!}-1)z^2+(-1\cdot\frac{1}{2!}+1+\frac{1}{4!})z^4+\ldots
   \textrm{ for } |z|<1.$$

  \item[(ii)]
  $$ \ $$
  $\begin{array}{l}
  \ds\frac{\cosh z}{1+z^2}=\frac{\cosh z}{(z+i)(z-i)}
  \end{array}
  \ \ \
  \begin{array}{c}
  \epsfig{file=272-1-9.eps, width=40mm}
  \end{array}$

 \begin{eqnarray*}
  &=&\frac{1}{z-i}\frac{1}{(z-i)+2i} \cosh z\\
  &=&\frac{1}{z-i}\frac{1}{2i}\frac{1}{1+\frac{z-i}{2i}}\cosh z\\
  &=&\frac{1}{z-i}\frac{1}{2i}\left[\textrm{ geometric series in
  }\frac{z-i}{2i}\textrm{ for }|z-i|<2\right]\\
  &&\left[\textrm{ Taylor series of }
  \cosh z \textrm{ about } z=i, \textrm{ for all } z\right]\\
  &=&\frac{1}{z-i}\frac{1}{2i}\left[1-\frac{z-i}{2i}+(\frac{z-i}{2i})^2-\ldots\right]\\
  &=&\left[\cosh
  (i) +\sinh (i)(z-i) +\frac {\cosh (i)}{2!}(z-i)^2+\ldots\right]\\
  &=& \frac{1}{2i}\frac{1}{z-i}\left[1-\frac{z-i}{2i}+\ldots\right]
  \left[\cos (1)+i  \sin (1)(z-i)+\ldots\right]\\
  &=&\frac{1}{2i}\left[\cos (1)(z-i)^{-1}+(-\frac{1}{2i}+i \sin
  (1))+\ldots\right]
  \end{eqnarray*}

  (Using $\cosh(ix)=\cos (x),\ \sinh (ix)=i\sin(x)$)

  Res$\ds=\frac{\cos (1)}{2i}$ This is a Laurent series for $0<|z-i|<2$

  \end{description}

  \end{description}


\end{document}