\documentclass[a4paper,12pt]{article}
\usepackage{epsfig}
\begin{document}

\parindent=0pt

QUESTION

Evaluate the following directly by parameterising the curves:

\begin{description}

\item[(a)]
$\int_\gamma |z|\,dz$

where $\gamma$ is

\begin{description}

\item[(i)]
 the straight line joining $z=-1-i$ to $z=1+i$,

\item[(ii)]
 the upper unit semicircle centred on O, joining $z=1$ to $z=-1$,

\item[(iii)]
 the lower unit semicircle centred on O, joining $z=-1$ to $z=1$.

\end{description}

\item[(b)]
$\int_\gamma z^\frac{1}{2}\,dz$

where $\gamma$ is

\begin{description}

\item[(i)]
 the straight line joining $z=-1$ to $z=i$,

\item[(ii)]
 the left unit semicircle centered on O, joining $z=i$ to $z=-i$.

\end{description}

\item[(c)]
$\int_\gamma \sinh z\,dz$

where $\gamma$ is the straight line joining $z=0$ to $z=i$.

\end{description}

\bigskip

ANSWER
 \begin{description}

  \item[(a)]

   \begin{description}

   \item[(i)]
   $$ \ $$
   $\begin{array}{l}
   z(t)=(1+i)t,\ \ -1\leq t \leq 1\\
   |z(t)|=\sqrt{2}|t|\ \ \frac{dz}{dt}=1+i
   \end{array}
   \ \ \
   \begin{array}{c}
   \setlength{\unitlength}{.5in}
\begin{picture}(4,3)
\put(0,1.5){\line(1,0){3}} \put(1.5,0){\line(0,1){3}}
\put(.5,1.4){\line(0,1){.2}} \put(2.5,1.4){\line(0,1){.2}}
\put(1.4,.5){\line(1,0){.2}} \put(1.4,2.5){\line(1,0){.2}}
\put(.5,.5){\line(1,1){2}} \put(.4,.4){$\bullet$}
\put(2.4,2.4){$\bullet$} \put(.4,1.2){$-1$} \put(2.4,1.2){1}
\put(1.7,2.5){$i$} \put(1.7,.5){$-i$}
\end{picture}
   \end{array}$

   \begin{eqnarray*}
   \int_\gamma |z|\,dz&=&\int_{-1}^1 \sqrt{2}|t|(1+i) \\
   &=&2\int_0^1 \sqrt{2}t(1+i)\,dt \\
   &=&2\sqrt{2}(1+i)\left[\frac{1}{2}t^2\right]_0^1 \\
   &=&(1+i)\sqrt{2}
   \end{eqnarray*}


   \item[(ii)]
   $$ \ $$
   $\begin{array}{l}
   z(t)=e^{it},\ \ 0 \leq t \leq \pi\\
   |z(t)|=1\ \ \frac{dz}{dt}=ie^{it}
   \end{array}
   \ \ \
   \begin{array}{c}
   \epsfig{file=272-1-1.eps, width=30mm}
   \end{array}$


   $$\int_\gamma |z|\,dz=\int_0^\pi
   ie^{it}\,dt=\left[e^{it}\right]_0^\pi=-2$$



   \item[(iii)]
   $$ \ $$
   $\begin{array}{l}
   z(t)=e^{it},\ \\ \pi \leq t \leq 2 \pi\\
   \int_\gamma |z|\,dz=[e^{it}]_\pi^{2\pi}=2
   \end{array}
   \ \ \
   \begin{array}{c}
   \epsfig{file=272-1-2.eps, width=30mm}
   \end{array}$

   \end{description}

  \item[(b)]

   \begin{description}

  \item[(i)]
   $$ \ $$
   $\begin{array}{l}
   z(t)=(1+i)t-1,\ \ 0 \leq t \leq 1\\
   \frac{dz}{dt}=1+i
   \end{array}
   \ \ \
   \begin{array}{c}
    \setlength{\unitlength}{.5in}
\begin{picture}(4,6)
\put(0,2){\line(1,0){3}} \put(2,1){\line(0,1){3}}
\put(.5,2){\line(1,1){1.5}} \put(.5,1.7){$-1$} \put(2.2,3.5){$i$}
\put(.4,1.9){$\bullet$} \put(1.9,3.4){$\bullet$}
\end{picture}
   \end{array}$

   \begin{eqnarray*}
    \int_\gamma z^{\frac{1}{2}}\,
    dz&=&\int_0^1\left[(1+i)t-1\right]^{\frac{1}{2}}(1+i)\,dt\\
    &=&\left[\frac{2}{3}((1+i)t-1)^{\frac{3}{2}}\right]\\
    &=&\frac{2}{3}\left[(i)^\frac{3}{2}-(-1)^\frac{3}{2}\right]\\
    &=&\frac{2}{3}\left(e^{\frac{3}{2}\frac{i \pi}{2}}-e^{\frac{3}{2}i
    \pi}\right)\\
    &=&\frac{2}{3}\left[\frac{1}{\sqrt{2}}(-1+i)-(-i)\right]\\
    &=&\frac{2}{3}\left[-\frac{1}{\sqrt{2}}+\left(1+\frac{1}{\sqrt{2}}\right)i\right]
   \end{eqnarray*}


   \item[(ii)]
   $$ \ $$
   $\begin{array}{l}
   z(t)=e^{it},\ \ \frac{\pi}{2} \leq t \leq \frac{3 \pi}{2}
   \end{array}
   \ \ \
   \begin{array}{c}
   \epsfig{file=272-1-3.eps, width=30mm}
   \end{array}$


    $$\int_\gamma
    z^\frac{1}{2}\,dz=\int_\frac{\pi}{2}^\frac{3\pi}{2}
    \left(e^{it}\right)^\frac{1}{2}ie^{it}\,dt
    =\left[\frac{2}{3}e^{i\frac{3}{2}t}\right]_\frac{\pi}{2}^\frac{3
    \pi}{2}$$
    $$=\frac{2}{3}\left(e^\frac{9 \pi i}{4}-e^\frac{3 \pi i}{4}\right)
    =\frac{2}{3}\frac{1}{\sqrt{2}}\left[(1+i)-(-1+i)\right]
    =\frac{2 \sqrt{2}}{3}$$

   \end{description}

  \item[(c)]
  $$ \ $$
  $\begin{array}{l}
  z(t)=it,\ \ 0 \leq t \leq 1,\ \ \frac{dz}{dt}=i
  \end{array}
  \ \ \
  \begin{array}{c}
 \setlength{\unitlength}{.5in}
 \begin{picture}(4,2)
\put(0,.5){\line(1,0){2}} \put(1,0){\line(0,1){2}}
\put(1,0){\vector(0,1){1}} \put(.92,.42){$\bullet$}
\put(.92,1.42){$\bullet$} \put(1.1,.2){0} \put(1.1,1.5){$i$}
 \end{picture}


  \end{array}$


  $$\int_\gamma \sinh z\, dz=\int_0^1 i \sinh it \, dt
  =\int_0^1 \frac{i}{2}(e^{it}-e^{-it})\,dt$$
  $$=\int_0^1-\sin t\, dt
  =\left[\cos t\right]_0^1
  =\cos (1) -1$$


  \end{description}


\end{document}