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\newcommand{\un}{\underline}
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\newcommand{\pl}{\partial}
\parindent=0pt
\begin{document}

{\bf Question}

Calculate $\arg(z)$ defined on $[0,2\pi)$ for the following values
of $z$:

$$(i)z=0,\ (ii)z=1+i,\ (iii)z=i,\ (iv)z=-1+i$$

$$(v)z=-1.\ (vi)z=-1-i,\ (vii)z=-i,\ (viii)z=1-i$$

\medskip

{\bf Answer}

$\log z=\log |z|+i\arg(z)\ \ 0 \leq \arg(z) < 2\pi$

\begin{description}
\item[(i)]
$\log(0)=\undb{\log|0|}+i\undb{\arg(0)}$

\ \ \ \ \ \ \ \ \ \ $-\infty$\ \ \ not defined

\item[(ii)]
$\log(1+i)=\log|1+i|+i\arg(1+i)=\un{\sqrt{2}+\ds\frac{i\pi}{4}}$

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\end{picture} \vspace{1in}

\item[(iii)]
$\log(i)=\log|i|+i\arg(i)=\un{0+\ds\frac{i\pi}{2}}$

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\item[(iv)]
$\log(-1+i)=\log|-1+i|+i\arg(-1+i)=\un{\sqrt{2}+\ds\frac{3i\pi}{4}}$

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\put(.5,.5){\circle*{.125}}

\put(1,0){\line(-1,1){.5}}
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\item[(v)]
$\log(-1)=\log|-1|+i\arg(-1)=0+i\pi=\un{i\pi}$

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\item[(vi)]
$\log(-1-i)=\log|-1-i|+i\arg(-1-i)=\un{\log\sqrt{2}+\ds\frac{5i\pi}{4}}$

NB Not $-\ds\frac{3\pi}{4}i$ as before since we're now defining
arg on

\ \ $0 \leq \arg(z) < 2\pi$

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\newpage
\item[(vii)]
$\log(-i)=\log|-i|+i\arg(-i)=0+i\ds\frac{3\pi}{2}=\un{\ds\frac{3i\pi}{2}}$

NB Not $-\ds\frac{i\pi}{2}$ as before since we're now defining arg
on

\ \ $0 \leq \arg(z) < 2\pi$

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\put(1,-.5){\circle*{.125}}

\end{picture} \vspace{1in}

\item[(viii)]
$\log(1-i)=\log|1-i|+i\arg(1-i)=\un{\log\sqrt{2}+\ds\frac{7i\pi}{4}}$

NB Not $-\ds\frac{i\pi}{4}$ as before since we're now defining arg
on

\ \ $0 \leq \arg(z) < 2\pi$

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\end{description}

\end{document}