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\begin{document}

{\bf Question}

Show that the formula of real analysis
$\rm{Log}(ab)=\rm{Log}(a)+\rm{Log}(b)$ where $a,\ b \not\in C$,
does not always hold in the full complex plane, i.e., when $a,\ b
\in C$. However, show that
$\rm{Log}(ab)=\rm{Log}(a)+\rm{Log}(b)+2n\pi i$ where $n$ is an
integer.


\medskip

{\bf Answer}

\begin{eqnarray*} \rm{Log} & = & \log|ab|+i\rm{Arg}(ab)\\ & = &
\log|a|+\log|b|+i\rm{Arg}(ab) \end{eqnarray*}

Now if $\rm{Arg}(ab)=\rm{Arg}(a)+\rm{Log}(b)\ (\star)$ we have

\begin{eqnarray*} \rm{Log}(ab) & = &
\log|a|+i\rm{Arg}(a)+\log|b|+i\rm{Arg}(b)\\ & = &
\rm{Log}(a)+\rm{Log}(b) \end{eqnarray*}

But this only works if $(\star)$ holds.

What happens if
$\left\{\begin{array}{l}\rm{Arg}(a)=\ds\frac{3\pi}{4}\\
\rm{Arg}(b)=\ds\frac{3\pi}{4} \end{array} \right\}$ say.

$$\rm{Arg}(a)+\rm{Arg}(b)=\ds\frac{3\pi}{4}+\ds\frac{3\pi}{4}=\ds\frac{3\pi}{2}$$

But $\ds\frac{3\pi}{2}$ doesn't lie in the range on \un{Arg}

since $-\pi < \rm{Arg} \leq \pi$ and $\pi<\ds\frac{3\pi}{2}$, so
here we need

$$\rm{Arg}(ab)=\rm{Arg}(a)+\rm{Arg}(b)-\undb{2\pi}$$

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ to get the RHS back into
$(-\pi,\pi]$

so in general

$$\rm{Arg}(ab)=\rm{Arg}(a)+\rm{Arg}(b)+2n\pi i$$

$n$ integer ($>$ or $<0$)

NB for $\rm{Arg}(a)=\rm{Arg}(b)=0\ \
\rm{Log}(ab)=\rm{Log}(a)+\rm{Log}(b)$ the usual \un{real} $a,\ b$
formula.
\end{document}