\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\un}{\underline}
\newcommand{\undb}{\underbrace}
\newcommand{\pl}{\partial}
\parindent=0pt
\begin{document}

{\bf Question}

Consider the functions

\begin{eqnarray*} g(z) & = & \rm{Log}(z),\\ h(z) & = & \rm{Log}(-z),\\ f(z)
& = & g(z)-h(z), \end{eqnarray*}

\begin{description}
\item[(i)]
Show that $g(z)$ is defined with $\arg(z)$ on $(-\pi,\pi]$ and
that $h(z)$ is defined with $\arg(z)$ on $(-2\pi,0]$. (Hint: note
the difference between $\\rm{Arg}$ and $\arg$).

\item[(ii)]
Draw the branch cut which makes the function $f(z)$ single valued
and continuous.

\item[(iii)]
Calculate $\ds\frac{df}{dz}$. What does this suggest about the
function $f(z)$?

\item[(iv)]
Calculate $f(i)$.

\item[(v)]
Calculate $f(-i)$.

\item[(vi)]
Do your answers to (ii) and (iii) contradict your answers to
(iii)? (Hint: Consider a general point with $Im(z)>0$ and then one
with $Im(z)<0$).
\end{description}

\medskip

{\bf Answer}

\begin{description}
\item[(i)]
$\begin{array}{r} g(z)\\ \rm{so} \end{array} \begin{array}{l}
\rm{Log}|z|+i\rm{Arg}(z)\\ \rm{Arg}(z) \in (-\pi,\pi]\ \rm{by\
definition} \end{array}$

$\begin{array} {rcl} h(z) & = & \log|-z|+i\rm{Arg}(-z)\\ & = &
\log|z|+i\rm{Arg}(-z) \end{array}$

so $\begin{array}{rcccl} -\pi & < & \arg(z) & \leq & \pi\\ -\pi &
< & \arg(ze^{i\pi}) & \leq & \pi\\ -\pi & < & \arg(z)+\pi & \leq &
\pi\\ -2\pi & < & \arg(z) & \leq & 0 \end{array}$

\newpage
\item[(ii)]
$f(z)=\log(z)-\rm{Log}(-z)$

The branch cut for Log$(z)$ is

PICTURE \vspace{1in}

The branch cut for Log$(-z)$ is

PICTURE \vspace{1in}

Putting them together:

PICTURE \vspace{1in}

\item[(iii)]
\begin{eqnarray*} \ds\frac{df}{dz} & = &
\ds\frac{d}{dz}[\rm{Log}(z)]-\ds\frac{d}{dz}[\rm{Log}(-z)]\\ & = &
\ds\frac{1}{z}-\ds\frac{(-1)}{(-z)}=\ds\frac{1}{z}-\ds\frac{1}{z}=0
\end{eqnarray*}

This suggests that \un{$f$=constant} for all $z$.

\item[(iv)]
\begin{eqnarray*} f(i) & = & \rm{Log}(i)-\rm{Log}(-i)\\ & = &
\log|i|+i\rm{Arg}(i)-\log|i|-i\rm{Arg}(-i)\\ & = &
i(\rm{Arg}(i)-\rm{Arg}(-i))\\ & = &
i\left(\ds\frac{\pi}{2}-\left(\ds\frac{-\pi}{2}\right)\right)\\ &
= & \un{+i\pi} \end{eqnarray*}

\item[(v)]
\begin{eqnarray*} f(-i) & = & \rm{Log}(-i)-\rm{Log}(i)\\ & = &
\log|-i|+i\rm{Arg}(-i)-\log|+i|-i\rm{Arg}(i)\\ & = &
i(\rm{Arg}(-i)-\rm{Arg}(i))\\ & = &
i\left(-\ds\frac{\pi}{2}-\ds\frac{\pi}{2}\right)\\ & = &
\un{-i\pi} \end{eqnarray*}

\item[(vi)]
At first sight (iv) $\ne$ (v) which contradicts (iii).

Resolution is: The cut in (ii) extends along all of the real axis
, \un{partitioning} $Im(z)>0$ from $Im(z)<0$. The derivative is
correst

but $\left\{\begin{array}{l} f=const(1)\ \rm{for}\ Im(z)>0\\
f=const(2)\ \rm{for}\ Im(z)<0 \end{array} \right.$

This can be seen for arbitrary points:

Take $z=a+ib$ with \un{$b>0$}.

PICTURE \vspace{1in}

\begin{eqnarray*} f(a+ib) & = &
\log|a+ib|+i\rm{Arg}(a+ib)\\ & & -\log|-a-ib|-i\rm{Arg}(-a-ib)\\ &
= & i[\rm{Arg}(a+ib)-\rm{Arg}(-a-ib)]\\ & = &
i[\theta-(-\pi+\theta)]\\ & = & \un{i\pi} \end{eqnarray*}

Now take $z=x+iy$ with \un{$y>0$}.

PICTURE \vspace{1in}

\newpage
\begin{eqnarray*} f(x+iy) & = &
\log|x+iy|+i\rm{Arg}(x+iy)\\ & & -\log|-x-iy|-i\rm{Arg}(-x-iy)\\ &
= & i[\rm{Arg}(x+iy)-\rm{Arg}(-x-iy)]\\ & = &
i[\alpha-(\pi-\alpha)]\\ & = & \un{-i\pi} \end{eqnarray*}



\end{description}
\end{document}