Find all $z\in\mathbb C$ such that $\text{Log}\left(z+\sqrt{z^2-1}\right)=-i\,\text{Log}\left(iz+\sqrt{1-z^2}\right)$.
(Here, I assume, that $\text{Log}\,z=\ln|z|+i\,\text{Arg}\,z$ with $\text{Arg}\,z\in(-\pi,\pi]$.) I did it by modifications of the RHS
$$-i\,\text{Log}\left(iz+\sqrt{1-z^2}\right)=-i\,\text{Log}\left(i\left(z-i\sqrt{1-z^2}\right)\right)=-i\left(\text{Log}\,(i)+\text{Log}\left(z+\sqrt{z^2-1}\right)\right)=\frac{\pi}{2}-i\,\text{Log}\left(z+\sqrt{z^2-1}\right)$$ which is not correct, but after some tedious computations I found $$z=\frac{1}{\sqrt{2}}\cosh\left(\frac{\pi}{4}\right)-\frac{i}{\sqrt{2}}\sinh\left(\frac{\pi}{4}\right)$$ The problem is, that $\bar z$ is also a root, which I lost in the process. So my questions are
- How to proceed in order to solve the equation without losing anything?
- Is it possible to simplify the process of solution using the fact that $\text{Log}\left(iz+\sqrt{1-z^2}\right)=\text{Argsinh}\,(iz)$ and the LHS is almost $\text{Argcosh}\,(z)$ ?
- What would change if we replace princial $\text{Log}$ with multi-valued $\log$ in the original equation?
Edit2: resolved on AoPS
Edit: Maybe I should have proceed in this way. $$-i\,\text{Log}\left(iz+\sqrt{1-z^2}\right)=-i\,\text{Log}\left(iz+\sqrt{(-1)\left(z^2-1\right)}\right)=-i\,\text{Log}\left(iz\pm i\sqrt{z^2-1}\right)=-i\,\left(\text{Log}(i)+\text{Log}\left(z\pm\sqrt{z^2-1}\right)\right)=\frac{\pi}{2}-i\,\text{Log}\left(z\pm\sqrt{z^2-1}\right)$$ Now, focus on the variant with $+$ and I get $$z_0=\frac{1}{\sqrt{2}}\cosh\left(\frac{\pi}{4}\right)-\frac{i}{\sqrt{2}}\sinh\left(\frac{\pi}{4}\right)$$ as before. For this particular $z_0$ we have $$\text{Arg}(-1)=\pi\quad\text{and}\quad\text{Arg}\left(z_0^2-1\right)\approx-1.981\quad\Longrightarrow\quad\sqrt{(-1)\left(z_0^2-1\right)}=\sqrt{-1}\sqrt{z_0^2-1}$$ similarly, $$\text{Arg}(i)=\frac{\pi}{2}\quad\text{and}\quad\text{Arg}\left(z_0\pm\sqrt{z_0^2-1}\right)=\mp\frac{\pi}{4}\quad\Longrightarrow\quad\text{Log}\left(iz_0\pm i\sqrt{z_0^2-1}\right)=\text{Log}(i)+\text{Log}\left(z_0\pm\sqrt{z_0^2-1}\right)$$ so, the (possibly incorrect) steps above should be now justified. This won't work for $\bar{z_0}$ though, because $\text{Arg}\left(\bar{z_0}^2-1\right)\approx+1.981$, but I believe it could be made by a diffrent choice of $\text{Arg}$, namely for $\text{Arg}\,w\in[-\pi,\pi)$.
But still, I'm missing the other root. WA claims that $$\text{Log}\left(z+\sqrt{z^2-1}\right)=\frac{\pi}{2}-i\,\text{Log}\left(z-\sqrt{z^2-1}\right)$$ has a root $\approx0.93664 + 0.614243 i$, which numerically corresponds with $\bar{z_0}$. I don't see how to solve this one though...