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:
$i\frac {c}{a} - \frac {b}{a} = \frac {(e^{-2ix} - 1)}{e^{-2ix} + 1}\\
\frac {ic - b}{a} = \frac {(e^{-ix} - e^{ix})}{e^{-ix} + e^{ix}}\\
\frac {-c - bi}{a} = \frac {i(e^{-ix} - e^{ix})}{e^{-ix} + e^{ix}}\\
\frac {-c - bi}{a} = \frac {e^{ix} - e^{-ix}}{i(e^{-ix} + e^{ix})}\\
\frac {-c - bi}{a} = \tan x\\
x = \tan^{-1} (-\frac {c}{a} - i\frac {b}{a})$
Assuming $a,c&gt;0$, if you write $a+ic = re^{i\theta}$ where $r=\sqrt{a^2+c^2}$ and $\theta=\tan^{-1}(\frac{c}{a})$, then
$$-a+ic = re^{i(\pi - \theta)}$$
Since both numbers are positive the argument will be $\pi - \theta$ for $-a+ic$.
Now put this in your expression to get:
$$\frac{1}{2}i\log{\frac{a+ic}{a-ic}}=\frac{1}{2}i\log{\frac{re^{i\theta}}{re^{i(\pi -\theta)}}} = \frac{1}{2}i\log e^{i(2\theta-\pi)}=\frac{1}{2}i\times i(2\theta - \pi)$$
which becomes
$$-\frac{1}{2}(2\theta - \pi)=\frac{\pi}{2}-\theta=\frac{\pi}{2}-\tan^{-1}\frac{c}{a}=\tan^{-1}\frac{a}{c}$$
Use the fact $$\tan^{-1}x+\tan^{-1}\frac{1}{x}=\frac{\pi}{2}$$ for all $x&gt;0$
.The expression might lead to different expressions depending on the values of $a,c$ I took the case where both are positive. It can be shown for rest of the cases as well. Hope this helps....