$\\Log$ denotes the multi-valued natural log defined over the nonzero complex numbers.
Assume $k$ is the particular choice of value of $\cosh^{-1}(z)$ for this post.
I use $=_c$ to denote that if you graph the difference of the two expressions on either side, the graph will be piecewise equal to a number of constants
Why doesn't $$\displaystyle\frac{\tanh\left(\frac{\cosh^{-1}(z)}{2}\right)+i}{\tanh\left(\frac{\cosh^{-1}(z)}{2}\right)-i}+C=_c-\frac{z}{\sqrt{1-{z^2}}+1}$$ $\tag1$ imply
$$\\Log\space\left[\frac{\tanh\left(\frac{\cosh^{-1}(z)}{2}\right)+i}{\tanh\left(\frac{\cosh^{-1}(z)}{2}\right)-i}+C\right]=_c\\Log\left(-\frac{z}{\sqrt{1-{z^2}}+1}\right)+i(2n\pi){ ?}$$ $\tag2$
This seems to break the rule that if $z=w$ then the set of values you can choose for $\\Log(z)$ are the same set set of values that you can choose for $\\Log(w)$
I know that equation (1) is true. This would seem to imply that equation (2) is true, However, equation (2) is not true. One can show that it is not true by differentiating both sides and seeing that their first derivatives do not match up. Therefore, it cannot be the case that equation (1) being true implies that equation (2) is true.
Why/how doesn't the fact that (1) is true imply that (2) is true?
You can see here that (1) is true and here that (2) is not true unless $C$ is zero
Best Answer
$\DeclareMathOperator{\arcosh}{arcosh}$ $\DeclareMathOperator{\arsech}{arsech}$ $\DeclareMathOperator{\sgn}{sgn}$ $\newcommand{\i}{\mathrm{i}}$
Work with the conventional principal branches of the square root, logarithm, and inverse trigonometric functions. Recall that if $z\in\mathbb{C}\setminus(-\infty,1]$, then $$\tanh (\tfrac{1}{2}\arcosh z)=\sqrt{\frac{z-1}{z+1}}=\frac{z}{z+1}\sqrt{1-z^{-2}}\text{.}$$
What's true is as follows: if $z\in\mathbb{C}\setminus \mathbb{R}$, then $$\boxed{\left(\frac{\i+\tfrac{z}{z+1}\sqrt{1-z^{-2}}}{\i-\tfrac{z}{z+1}\sqrt{1-z^{-2}}}\right)^{\sgn\Im z}=\frac{1+\sqrt{1-z^2}}{z}}\text{;}$$ equivalently, $$\frac{\i+\tfrac{z}{z+1}\sqrt{1-z^{-2}}}{\i-\tfrac{z}{z+1}\sqrt{1-z^{-2}}}=\frac{1+(\sgn\Im z)\sqrt{1-z^2}}{z}\text{.}$$ You can prove this equality with elementary algebra if you use the fact that, when $z\in\mathbb{C}\setminus \mathbb{R}$, $$z\sqrt{1-z^{-2}}=\i(\sgn\Im z)\sqrt{1-z^2}\text{.}$$ Then since the inverse tangent and inverse hyperbolic secant are related to the logarithm by $$\begin{align} \arctan z &=\tfrac{1}{2\i}\log \frac{\i + z}{\i - z} & z/\i &\in\mathbb{C}\setminus((-\infty,-1]\cup[1,\infty))\\ \arsech z &= \log\left(\frac{1+\sqrt{1-z^2}}{z}\right) & z &\in\mathbb{C}\setminus((-\infty,0]\cup[1,\infty))\text{,} \end{align}$$ taking the logarithm of both sides of the boxed equality yields $$\arsech z = -2\i(\sgn \Im z)\arctan \left(\tanh \tfrac{1}{2}(\arcosh z)\right)\text{.}$$ Note that many of the identities are given on $z\in\mathbb{C}\setminus \mathbb{R}$, i.e., away from the real axis. Every point on the real axis lies on the branch cut of the left- or right-hand sides of these identities—whether with Mathematica or otherwise, you can't reasonably plot the functions involved without understanding the limit interpretation the software is taking at the cut.