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\begin{document}

{\bf Question}

For each of the following functions, give a domain on which a
continuous branch can be defined.

$(i)\rm{Log}(1+z),\ (ii)\log(1+z),\ (iii)\rm{Log}(1+z^2),\
(iv)(z-1)^{\frac{1}{3}}\ (v)(z^2-1)^{\frac{1}{3}}.$


\medskip

{\bf Answer}

\begin{description}
\item[(i)]
\un{L}og$(1+z)$ has \rm{Arg}$(1+z)$ i.e., $-\pi<\arg(1+z) \leq
\pi$

so we need a branch cut from $z=-1$

PICTURE \vspace{2in}

\item[(ii)]
$\log(1+z)=\log|1+z|+i\undb{\arg(1+z)}$

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ what arg
though?

We need to define the branch of arg. Let's \un{choose} $0 <
\arg(1+z) \leq 2pi$, then we have a branch point at $z=-1$ and a
cut between $z=-1$ and $\un{+\infty}$.

PICTURE \vspace{2in}

\newpage
\item[(iii)]
Log$(1+z^2)=\log|1+z^2|+i$Arg$(1+z^2)$

Branch points where $1+z^2=0 \Rightarrow z=\pm i$

Now for \un{Arg} we have a cut where

PICTURE \vspace{2in}

\item[(iv)]
$(z-1)^{\frac{1}{3}}=e^{\frac{1}{3}\log(z-1)}=e^{\frac{1}{3}\log|z-1|+\frac{i\arg(z-1)}{3}}$

But again, we need to choose a branch. If we choose \un{A}rg we
have $-\pi < \arg(z-1) \leq \pi$ and we need a cut when $(z-1)<0$,
i.e.,

PICTURE \vspace{2in}

If we choose $0<\arg(z)\leq 2pi$ we need a cut when $z-1>0$, i.e.,

PICTURE \vspace{2in}

\item[(v)]
$(z^2-1)^{\frac{1}{3}}$ has branch points where

$z^2-1=0 \Rightarrow z=\pm 1$

Again we have to choose the branch. If \un{A}rg we need cut where
$(z^2-1)<0$.

PICTURE \vspace{2in}

If we choose $0<\arg(z^2-1)\leq 2\pi$ then we need a cut where
$(z^2-1)>0$

PICTURE \vspace{1in}

(Check this using the log-definitions.)
\end{description}

\end{document}