Logarithm to trigonometry conversion I have an expression like
$$\frac{1}{2} i \log \left(\frac{a-b+i c}{a+b-i c}\right)$$
I was wondering if it is possible to write it in terms of trig functions. I guess it is possible to write all logs in some form of trig due to Euler's relation.
As an attempt to solve the problem, I was trying to play around the relation
$$\frac{1}{2} i \log \left(\frac{a+i c}{-a+i c}\right)=\tan^{-1}\left(\frac{a}{c}\right)$$ to change into arctan function but does not look like this alone is going to help in this case.
Any suggestions?

$\frac {i}{2}\ln \frac {a+b+ic}{a-b-ic} = x\\
\ln \frac {a+b+ic}{a-b-ic} = -2ix\\
\frac {a+b+ic}{a-b-ic} = e^{-2ix}\\
a+b+ic = e^{-2ix}(a-b-ic)\\
a+b+ic = e^{-2ix}(a-b-ic)\\
(ic-b)(1+e^{-2ix}) = a(e^{-2ix} - 1)\\
\frac {ic - b}{a} = \frac {(e^{-2ix} - 1)}{e^{-2ix} + 1}$
At this point the right hand side equals very nearly $\tan x.$  That is:
$i\frac {c}{a} - \frac {b}{a} = \frac {(e^{-2ix} - 1)}{e^{-2ix} + 1}\\
\frac {ic - b}{a} = \frac {(e^{-ix} - e^{ix})}{e^{-ix} + e^{ix}}\\
\frac {-c - bi}{a} = \frac {i(e^{-ix} - e^{ix})}{e^{-ix} + e^{ix}}\\
\frac {-c - bi}{a} = \frac {e^{ix} - e^{-ix}}{i(e^{-ix} + e^{ix})}\\
\frac {-c - bi}{a} = \tan x\\
x = \tan^{-1} (-\frac {c}{a} - i\frac {b}{a})$
