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

{\bf Question}

Calculate $\rm{Log}(z)$ 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}

$\rm{Log}z=\log |z|+i\rm{Arg}(z)\ \ -\pi < \rm{Arg}z \leq \pi$

\begin{description}
\item[(i)]
$\rm{Log}0=\undb{\log 0}+i\undb{\rm{Arg}(0)}$

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

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

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\put(1.5,.5){\circle*{.125}}

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\end{picture} \vspace{1in}

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

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

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\item[(v)]
$\rm{Log}(-1)=\log|-1|+i\rm{Arg}(-1)=0+i\pi=\un{i\pi}$

NB $+\pi$ not $-\pi$ since \un{A}rg.

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\item[(vi)]
$\rm{Log}(-1-i)=\log|-1-i|+i\rm{Arg}(-1-i)
=\log\sqrt{2}+i\left(-\ds\frac{3\pi}{4}\right)
=\un{\log\sqrt{2}-\ds\frac{3i\pi}{4}}$

NB $\ds\frac{-3\pi}{4}$ not $\ds\frac{5\pi}{4}$ since $-\pi <
\rm{Arg} \leq \pi$

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\item[(vii)]
$\rm{Log}(-i)=\log|-i|+i\rm{Arg}(-i)=0-\ds\frac{i\pi}{2}=\un{-\ds\frac{i\pi}{2}}$

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\item[(viii)]
$\rm{Log}(1-i)=\log|1-i|+i\rm{Arg}(1-i)
=\log\sqrt{2}+i\left(-\ds\frac{\pi}{4}\right)
=\un{\log\sqrt{2}-\ds\frac{i\pi}{4}}$

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\end{description}
\end{document}