(1)

\begin{align*} \sin^{\bullet}\sin z & =-i\Log\left(i\sin z+\sqrt{1-\sin^{2}z}\right)\\ & =-i\Log\left(i\sin z+\sqrt{\cos^{2}z}\right)\\ & =-i\Log\left(i\sin z+\sqrt{\sgn^{2}\left(\cos z\right)}\left|\cos z\right|\right)\\ & =-i\Log\left(i\sin z\pm\sgn\left(\cos z\right)\left|\cos z\right|\right)\\ & =-i\Log\left(i\sin z\pm\cos z\right)\\ & =-i\Log\pm e^{\pm iz}\\ & =-i\Log e^{i\frac{1\mp1}{2}\pi}e^{\mp\Im z\pm i\Re z}\\ & =-i\Log e^{\mp\Im z}e^{i\left(\frac{1\mp1}{2}\pi\pm\Re z\right)}\\ & =-i\left\{ \mp\Im z+i\mod\left(\frac{1\mp1}{2}\pi\pm\Re z,-2\pi,\pi\right)\right\} \\ & =\mod\left(\frac{1\mp1}{2}\pi\pm\Re z,-2\pi,\pi\right)+i\Im z \end{align*} となるので\(\sqrt{\sgn^{2}\left(\cos z\right)}=\sgn\left(\cos z\right)\land-\pi<\Re z\leq\pi\)のとき恒等写像になる。
\begin{align*} & \sqrt{\sgn^{2}\left(\cos z\right)}=\sgn\left(\cos z\right)\land-\pi<\Re z\leq\pi\\ \Leftrightarrow & \left(-\frac{\pi}{2}<\Arg\cos z\leq\frac{\pi}{2}\lor\cos z=0\right)\land-\pi<\Re z\leq\pi\\ \Leftrightarrow & \left(-\frac{\pi}{2}<\Arg\cos z<\frac{\pi}{2}\lor\Arg\cos z=\frac{\pi}{2}\lor\cos z=0\right)\land-\pi<\Re z\leq\pi\\ \Leftrightarrow & \left(0<\Re\cos z\lor\left(\Re\cos z=0\land0<\Im\cos z\right)\lor\cos z=0\right)\land-\pi<\Re z\leq\pi\\ \Leftrightarrow & \left(0<\Re\cos z\lor\left(\Re\cos z=0\land0\leq\Im\cos z\right)\right)\land-\pi<\Re z\leq\pi\\ \Leftrightarrow & \left(0<\cos\left(\Re z\right)\cosh\left(\Im z\right)\lor\left(\cos\left(\Re z\right)\cosh\left(\Im z\right)=0\land0\leq-\sin\left(\Re z\right)\sinh\left(\Im z\right)\right)\right)\land-\pi<\Re z\leq\pi\\ \Leftrightarrow & \left(0<\cos\left(\Re z\right)\lor\left(\cos\left(\Re z\right)=0\land\sin\left(\Re z\right)\sinh\left(\Im z\right)\leq0\right)\right)\land-\pi<\Re z\leq\pi\\ \Leftrightarrow & -\frac{\pi}{2}<\Re z<\frac{\pi}{2}\lor\left(\Re z=-\frac{\pi}{2}\land0\leq\Im z\right)\lor\left(\Re z=\frac{\pi}{2}\land\Im z\leq0\right) \end{align*} これより題意は成り立つ。

(2)

\begin{align*} \cos^{\bullet}\cos z & =-i\Log\left(\cos z+i\sqrt{1-\cos^{2}z}\right)\\ & =-i\Log\left(\cos z+i\sqrt{\sin^{2}z}\right)\\ & =-i\Log\left(\cos z+i\sqrt{\sgn^{2}\left(\sin z\right)}\left|\sin z\right|\right)\\ & =-i\Log\left(\cos z\pm i\sgn\left(\sin z\right)\left|\sin z\right|\right)\\ & =-i\Log\left(\cos z\pm i\sin z\right)\\ & =-i\Log e^{\pm iz}\\ & =-i\Log e^{\mp\Im z\pm i\Re z}\\ & =-i\left\{ \mp\Im z+i\mod\left(\pm\Re z,-2\pi,\pi\right)\right\} \\ & =\mod\left(\pm\Re z,-2\pi,\pi\right)\pm i\Im z \end{align*} となるので\(\sqrt{\sgn^{2}\left(\sin z\right)}=\sgn\left(\sin z\right)\land-\pi<\Re z\leq\pi\)のとき恒等写像になる。
\begin{align*} & \sqrt{\sgn^{2}\left(\sin z\right)}=\sgn\left(\sin z\right)\land-\pi<\Re z\leq\pi\\ \Leftrightarrow & \left(-\frac{\pi}{2}<\Arg\sin z\leq\frac{\pi}{2}\lor\sin z=0\right)\land-\pi<\Re z\leq\pi\\ \Leftrightarrow & \left(-\frac{\pi}{2}<\Arg\sin z<\frac{\pi}{2}\lor\Arg\sin z=\frac{\pi}{2}\lor\sin z=0\right)\land-\pi<\Re z\leq\pi\\ \Leftrightarrow & \left(0<\Re\sin z\lor\left(\Re\sin z=0\land0<\Im\sin z\right)\lor\sin z=0\right)\land-\pi<\Re z\leq\pi\\ \Leftrightarrow & \left(0<\Re\sin z\lor\left(\Re\sin z=0\land0\leq\Im\sin z\right)\right)\land-\pi<\Re z\leq\pi\\ \Leftrightarrow & \left(0<\sin\left(\Re z\right)\cosh\left(\Im z\right)\lor\left(\sin\left(\Re z\right)\cosh\left(\Im z\right)=0\land0\leq\Im\cos\left(\Re z\right)\sinh\left(\Im z\right)\right)\right)\land-\pi<\Re z\leq\pi\\ \Leftrightarrow & \left(0<\sin\left(\Re z\right)\lor\left(\sin\left(\Re z\right)=0\land0\leq\Im\cos\left(\Re z\right)\sinh\left(\Im z\right)\right)\right)\land-\pi<\Re z\leq\pi\\ \Leftrightarrow & 0<\Re z<\pi\lor\left(\Re z=0\land0\leq\Im z\right)\lor\left(\Re z=\pi\land\Im z\leq0\right) \end{align*} これより題意は成り立つ。

(3)

\begin{align*} \tan^{\bullet}\tan z & =\frac{1}{2}i\left\{ \Log\left(1-i\tan z\right)-\Log\left(1+i\tan z\right)\right\} \\ & =\frac{1}{2}i\left\{ \ln\left|1-i\tan z\right|+i\Arg\left(1-i\tan z\right)-\ln\left|1+i\tan z\right|-i\Arg\left(1+i\tan z\right)\right\} \\ & =\frac{1}{2}i\left\{ \ln\left|\frac{1-i\tan z}{1+i\tan z}\right|+i\left(\Arg\left(1-i\tan z\right)-\Arg\left(1+i\tan z\right)\right)\right\} \\ & =\frac{1}{2}i\left\{ \ln\left|\frac{1-i\tan z}{1+i\tan z}\right|+i\left(\Arg\left(\frac{1-i\tan z}{1+i\tan z}\right)+2\pi H_{0}\left(\Arg\left(1-i\tan z\right)+\Arg\left(1+i\tan z\right)^{-1}-\pi\right)-2\pi H_{1}\left(-\pi-\Arg\left(1-i\tan z\right)-\Arg\left(1+i\tan z\right)^{-1}\right)-2\pi\delta_{\pi,\Arg\left(1+i\tan z\right)}\right)\right\} \\ & =\frac{1}{2}i\left\{ \ln\left|e^{-2iz}\right|+i\left(\Arg\left(e^{-2iz}\right)+2\pi H_{0}\left(\Arg\left(1-i\tan z\right)+\Arg\left(1-i\tan\overline{z}\right)-\pi\right)-2\pi H_{1}\left(-\pi-\Arg\left(1-i\tan z\right)-\Arg\left(1-i\tan\overline{z}\right)\right)-2\pi\delta_{\pi,\Arg\left(1+i\tan z\right)}\right)\right\} \\ & =\frac{1}{2}i\left\{ \ln\left|e^{-2i\Re z}e^{2\Im z}\right|+i\left(\Arg\left(e^{-2i\Re z}e^{2\Im z}\right)+2\pi H_{0}\left(\Arg\left(e^{2\Im z}+e^{-2i\Re z}\right)+\Arg\left(e^{-2\Im z}+e^{-2i\Re z}\right)-\pi\right)-2\pi H_{1}\left(-\pi-\Arg\left(e^{2\Im z}+e^{-2i\Re z}\right)-\Arg\left(e^{-2\Im z}+e^{-2i\Re z}\right)\right)-2\pi\delta_{\pi,\Arg\left(\frac{1}{\cos\left(2\Re z\right)+\cosh\left(2\Im z\right)}\left(e^{-2\Im z}+e^{2i\Re z}\right)\right)}\right)\right\} \\ & =\frac{1}{2}i\left\{ \ln e^{2\Im z}+i\left(\Arg\left(e^{-2i\Re z}\right)-2\pi T\left(\Re z\in\pi\left(\frac{1}{2}+\mathbb{Z}\right)\land\Im z>0\right)\right)\right\} \\ & =\frac{1}{2}i\left\{ 2\Im z+i\left(\mod\left(-2\Re z,-2\pi,\pi\right)-2\pi T\left(\Re z\in\pi\left(\frac{1}{2}+\mathbb{Z}\right)\land\Im z>0\right)\right)\right\} \\ & =\frac{1}{2}i\left\{ 2\Im z+i\left(-2\mod\left(\Re z,\pi,-\frac{\pi}{2}\right)-2\pi T\left(\Re z\in\pi\left(\frac{1}{2}+\mathbb{Z}\right)\right)T\left(\Im z>0\right)\right)\right\} \\ & =\mod\left(\Re z,\pi,-\frac{\pi}{2}\right)+\pi T\left(\Re z\in\pi\left(\frac{1}{2}+\mathbb{Z}\right)\right)T\left(\Im z>0\right)+i\Im z \end{align*} これより、\(\tan z\)は\(z=\pm\frac{\pi}{2}\)のとき定義できないので、\(-\frac{\pi}{2}<\Re z<\frac{\pi}{2}\lor\left(\Re z=-\frac{\pi}{2}\land\Im z<0\right)\lor\left(\Re z=\frac{\pi}{2}\land\Im z>0\right)\)のとき恒等写像になる。

(4)

\begin{align*} \sin^{-1,\bullet}\sin^{-1}z & =\sin^{\bullet}\frac{1}{\sin^{-1}z}\\ & =\sin^{\bullet}\sin z \end{align*} となるので、\(\sin^{\bullet}\sin z\)が恒等写像になる領域と同じになるが、\(\sin^{-1}z\)が定義できない領域を除くと、
\begin{align*} & z\ne0\land\left(-\frac{\pi}{2}<\Re z<\frac{\pi}{2}\lor\left(\Re z=-\frac{\pi}{2}\land0\leq\Im z\right)\lor\left(\Re z=\frac{\pi}{2}\land\Im z\leq0\right)\right)\\ \Leftrightarrow & \left(z\ne0\land-\frac{\pi}{2}<\Re z<\frac{\pi}{2}\right)\lor\left(\Re z=-\frac{\pi}{2}\land0\leq\Im z\right)\lor\left(\Re z=\frac{\pi}{2}\land\Im z\leq0\right) \end{align*} となる。

(5)

\begin{align*} \cos^{-1,\bullet}\cos^{-1}z & =\cos^{\bullet}\frac{1}{\cos^{-1}z}\\ & =\cos^{\bullet}\cos z \end{align*} となるので、\(\cos^{\bullet}\cos z\)が恒等写像になる領域と同じになるが、\(\cos^{-1}z\)が定義できない領域を除くと、
\begin{align*} & z\ne\frac{\pi}{2}\land\left(0<\Re z<\pi\lor\left(\Re z=0\land0\leq\Im z\right)\lor\left(\Re z=\pi\land\Im z\leq0\right)\right)\\ \Leftrightarrow & \left(z\ne\frac{\pi}{2}\land0<\Re z<\pi\right)\lor\left(\Re z=0\land0\leq\Im z\right)\lor\left(\Re z=\pi\land\Im z\leq0\right) \end{align*} となる。

(6)

\begin{align*} \tan^{-1,\bullet}\tan^{-1}z & =\tan^{\bullet}\frac{1}{\tan^{-1}z}\\ & =\tan^{\bullet}\tan z \end{align*} となるので、\(\tan^{\bullet}\tan z\)が恒等写像になる領域と同じになるが、\(\tan^{-1}z\)が定義できない領域を除くと、
\begin{align*} & z\ne0\land\left(-\frac{\pi}{2}<\Re z<\frac{\pi}{2}\lor\left(\Re z=-\frac{\pi}{2}\land\Im z<0\right)\lor\left(\Re z=\frac{\pi}{2}\land\Im z>0\right)\right)\\ \Leftrightarrow & \left(z\ne0\land-\frac{\pi}{2}<\Re z<\frac{\pi}{2}\right)\lor\left(\Re z=-\frac{\pi}{2}\land\Im z<0\right)\lor\left(\Re z=\frac{\pi}{2}\land\Im z>0\right) \end{align*} となる。

(1)

\begin{align*} \sinh^{\bullet}\sinh z & =\sinh^{\bullet}\left(-ii\sinh z\right)\\ & =\sinh^{\bullet}\left(-i\sin\left(iz\right)\right)\\ & =i\sin^{\bullet}\left(-\sin\left(iz\right)\right)\\ & =-i\sin^{\bullet}\left(\sin\left(iz\right)\right) \end{align*} これより\(\sin^{\bullet}\left(\sin\left(iz\right)\right)=iz\)となるとき、恒等写像となるので、\(-\frac{\pi}{2}<\Re\left(iz\right)<\frac{\pi}{2}\lor\left(\Re\left(iz\right)=-\frac{\pi}{2}\land0\leq\Im\left(iz\right)\right)\lor\left(\Re\left(iz\right)=\frac{\pi}{2}\land\Im\left(iz\right)\leq0\right)\)すなわち、\(-\frac{\pi}{2}<\Im z<\frac{\pi}{2}\lor\left(\Im z=-\frac{\pi}{2}\land\Re z\leq0\right)\lor\left(\Im z=\frac{\pi}{2}\land0\leq\Re z\right)\)となる。

(2)

\begin{align*} \cosh^{\bullet}\cosh z & =\Log\left(\cosh z+\sqrt{\cosh z-1}\sqrt{\cosh z+1}\right)\\ & =\Log\left(\cosh z\pm_{1}\sqrt{\cosh^{2}z-1}\right)\\ & =\Log\left(\cosh z\pm_{1}\sqrt{\sinh^{2}z}\right)\\ & =\Log\left(\cosh z\pm_{1}\pm_{2}\sinh z\right)\\ & =\Log e^{\pm_{1}\pm_{2}z}\\ & =\pm_{1}\pm_{2}\Re z+i\left(\mod\left(\pm_{1}\pm_{2}\Im z,-2\pi,\pi\right)\right) \end{align*} これより恒等写像になるのは
\[ \sqrt{\cosh z-1}\sqrt{\cosh z+1}=\sqrt{\cosh^{2}z-1}\land\sqrt{\sinh^{2}z}=\sinh z\land-\pi<\Im z\leq\pi \] または
\[ \sqrt{\cosh z-1}\sqrt{\cosh z+1}=-\sqrt{\cosh^{2}z-1}\land\sqrt{\sinh^{2}z}=-\sinh z\land-\pi<\Im z\leq\pi \] となる。
\begin{align*} & \sqrt{\cosh z-1}\sqrt{\cosh z+1}=\sqrt{\cosh^{2}z-1}\land\sqrt{\sinh^{2}z}=\sinh z\land-\pi<\Im z\leq\pi\\ \Leftrightarrow & \left(-\frac{\pi}{2}<\Arg\cosh z\leq\frac{\pi}{2}\lor\left(\Arg\cosh z=\pi\land-1\leq\cosh z\right)\lor\cosh z=0\right)\land\left(-\frac{\pi}{2}<\Arg\sinh z\leq\frac{\pi}{2}\lor\sinh z=0\right)\land-\pi<\Im z\leq\pi\\ \Leftrightarrow & \left(-\frac{\pi}{2}<\Arg\cosh z<\frac{\pi}{2}\lor\Arg\cosh z=\frac{\pi}{2}\lor\left(\Arg\cosh z=\pi\land-1\leq\cosh z\right)\lor\cosh z=0\right)\land\left(-\frac{\pi}{2}<\Arg\sinh z<\frac{\pi}{2}\lor\Arg\sinh z=\frac{\pi}{2}\lor\sinh z=0\right)\land-\pi<\Im z\leq\pi\\ \Leftrightarrow & \left(0<\Re\cosh z\lor\left(\Re\cosh z=0\land0<\Im\cosh z\right)\lor\left(\Re\cosh z<0\land\Im\cosh z=0\land-1\leq\cosh z<0\right)\lor\cosh z=0\right)\land\left(0<\Re\sinh z\lor\left(\Re\sinh z=0\land0<\Im\sinh z\right)\lor\sinh z=0\right)\land-\pi<\Im z\leq\pi\\ \Leftrightarrow & \left(0<\Re\cosh z\lor\left(\Re\cosh z=0\land0\leq\Im\cosh z\right)\lor\left(\Im\cosh z=0\land-1\leq\cosh z<0\right)\right)\land\left(0<\Re\sinh z\lor\left(\Re\sinh z=0\land0\leq\Im\sinh z\right)\right)\land-\pi<\Im z\leq\pi\\ \Leftrightarrow & \left(0<\cos\left(\Im z\right)\lor\left(\cos\left(\Im z\right)=0\land0\leq\sinh\left(\Re z\right)\sin\left(\Im z\right)\right)\lor\left(\sinh\left(\Re z\right)\sin\left(\Im z\right)=0\land-\cosh^{-1}\left(\Re z\right)\leq\cos\left(\Im z\right)<0\right)\right)\land\left(0<\sinh\left(\Re z\right)\cos\left(\Im z\right)\lor\left(\sinh\left(\Re z\right)\cos\left(\Im z\right)=0\land0\leq\sin\left(\Im z\right)\right)\right)\land-\pi<\Im z\leq\pi\\ \Leftrightarrow & \left(\left(0<\cos\left(\Im z\right)\land0<\sinh\left(\Re z\right)\right)\lor\left(\sinh\left(\Re z\right)=0\land0\leq\sin\left(\Im z\right)\right)\lor\left(\cos\left(\Im z\right)=0\land0\leq\sin\left(\Im z\right)\land0\leq\sinh\left(\Re z\right)\right)\lor\left(\sin\left(\Im z\right)=0\land-\cosh^{-1}\left(\Re z\right)\leq\cos\left(\Im z\right)<0\land0<\sinh\left(\Re z\right)\cos\left(\Im z\right)\right)\lor\left(\sinh\left(\Re z\right)=0\land-\cosh^{-1}\left(\Re z\right)\leq\cos\left(\Im z\right)<0\land0\leq\sin\left(\Im z\right)\right)\right)\land-\pi<\Im z\leq\pi\\ \Leftrightarrow & \left(\left(0<\Re z\land-\frac{\pi}{2}<\Im z<\frac{\pi}{2}\right)\lor\left(\Re z=0\land0\leq\Im z\leq\pi\right)\lor\left(0\leq\Re z\land\Im z=\frac{\pi}{2}\right)\lor\left(\Re z=0\land\frac{\pi}{2}<\Im z\leq\pi\right)\right)\land-\pi<\Im z\leq\pi\\ \Leftrightarrow & \left(\left(0<\Re z\land-\frac{\pi}{2}<\Im z\leq\frac{\pi}{2}\right)\lor\left(\Re z=0\land0\leq\Im z\leq\pi\right)\right) \end{align*} \begin{align*} & \sqrt{\cosh z-1}\sqrt{\cosh z+1}=-\sqrt{\cosh^{2}z-1}\land\sqrt{\sinh^{2}z}=-\sinh z\land-\pi<\Im z\leq\pi\\ \Leftrightarrow & \left(\Arg\cosh z\leq-\frac{\pi}{2}\lor\frac{\pi}{2}<\Arg\cosh z<\pi\lor\left(\Arg\cosh z=\pi\land\cosh z\leq-1\right)\right)\land\left(\Arg\sinh z\leq-\frac{\pi}{2}\lor\left(\frac{\pi}{2}<\Arg\sinh z\lor\sinh z=0\right)\right)\land-\pi<\Im z\leq\pi\\ \Leftrightarrow & \left(\Arg\cosh z<-\frac{\pi}{2}\lor\frac{\pi}{2}<\Arg\cosh z<\pi\lor\Arg\cosh z=-\frac{\pi}{2}\lor\left(\Arg\cosh z=\pi\land\cosh z\leq-1\right)\right)\land\left(\Arg\sinh z<-\frac{\pi}{2}\lor\frac{\pi}{2}<\Arg\sinh z\lor\Arg\sinh z=-\frac{\pi}{2}\lor\sinh z=0\right)\land-\pi<\Im z\leq\pi\\ \Leftrightarrow & \left(\left(\Re\cosh z<0\land\Im\cosh z\ne0\right)\lor\left(\Re\cosh z=0\land\Im\cosh z<0\right)\lor\left(\Re\cosh z<0\land\Im\cosh z=0\land\cosh z\leq-1\right)\right)\land\left(\Re\sinh z<0\lor\left(\Re\sinh z=0\land\Im\sinh z\leq0\right)\right)\land-\pi<\Im z\leq\pi\\ \Leftrightarrow & \left(\left(\cosh\left(\Re z\right)\cos\left(\Im z\right)<0\land\sinh\left(\Re z\right)\sin\left(\Im z\right)\ne0\right)\lor\left(\cosh\left(\Re z\right)\cos\left(\Im z\right)=0\land\sinh\left(\Re z\right)\sin\left(\Im z\right)<0\right)\lor\left(\sinh\left(\Re z\right)\sin\left(\Im z\right)=0\land\cosh\left(\Re z\right)\cos\left(\Im z\right)\leq-1\right)\right)\land\left(\sinh\left(\Re z\right)\cos\left(\Im z\right)<0\lor\left(\sinh\left(\Re z\right)\cos\left(\Im z\right)=0\land\cosh\left(\Re z\right)\sin\left(\Im z\right)\leq0\right)\right)\land-\pi<\Im z\leq\pi\\ \Leftrightarrow & \left(\left(\cos\left(\Im z\right)<0\land\sinh\left(\Re z\right)\sin\left(\Im z\right)\ne0\right)\lor\left(\cos\left(\Im z\right)=0\land\sinh\left(\Re z\right)\sin\left(\Im z\right)<0\right)\lor\left(\sinh\left(\Re z\right)\sin\left(\Im z\right)=0\land\cosh\left(\Re z\right)\cos\left(\Im z\right)\leq-1\right)\right)\land\left(\sinh\left(\Re z\right)\cos\left(\Im z\right)<0\lor\left(\sinh\left(\Re z\right)\cos\left(\Im z\right)=0\land\sin\left(\Im z\right)\leq0\right)\right)\land-\pi<\Im z\leq\pi\\ \Leftrightarrow & \left(\left(\cos\left(\Im z\right)<0\land\sin\left(\Im z\right)\ne0\land0<\sinh\left(\Re z\right)\right)\lor\left(\cos\left(\Im z\right)=0\land\sinh\left(\Re z\right)\sin\left(\Im z\right)<0\land\sin\left(\Im z\right)<0\right)\lor\left(\sin\left(\Im z\right)=0\land\cos\left(\Im z\right)\leq-\cosh^{-1}\left(\Re z\right)\land\sinh\left(\Re z\right)\cos\left(\Im z\right)<0\right)\lor\left(\sinh\left(\Re z\right)=0\land\cos\left(\Im z\right)\leq-\cosh^{-1}\left(\Re z\right)\land\sin\left(\Im z\right)\leq0\right)\right)\land-\pi<\Im z\leq\pi\\ \Leftrightarrow & \left(\left(\cos\left(\Im z\right)<0\land\sin\left(\Im z\right)\ne0\land0<\sinh\left(\Re z\right)\right)\lor\left(\cos\left(\Im z\right)=0\land0<\sinh\left(\Re z\right)\land\sin\left(\Im z\right)<0\right)\lor\left(\sin\left(\Im z\right)=0\land\cos\left(\Im z\right)\leq-\cosh^{-1}\left(\Re z\right)\land\sinh\left(\Re z\right)\cos\left(\Im z\right)<0\right)\lor\left(\sinh\left(\Re z\right)=0\land\cos\left(\Im z\right)=-1\land\sin\left(\Im z\right)=0\right)\right)\land-\pi<\Im z\leq\pi\\ \Leftrightarrow & \left(\left(0<\Re z\land\left(\Im z<-\frac{\pi}{2}\lor\frac{\pi}{2}<\Im z\right)\right)\lor\left(0<\Re z\land\Im z=-\frac{\pi}{2}\right)\lor\left(0<\Re z\land\Im z=\pi\right)\lor\left(\Re z=0\land\Im z=\pi\right)\right)\land-\pi<\Im z\leq\pi\\ \Leftrightarrow & \left(0<\Re z\land\left(-\pi<\Im z\leq-\frac{\pi}{2}\lor\frac{\pi}{2}<\Im z\leq\pi\right)\right)\lor\left(\Re z=0\land\Im z=\pi\right) \end{align*} これより、
\begin{align*} & \left(\sqrt{\cosh z-1}\sqrt{\cosh z+1}=\sqrt{\cosh^{2}z-1}\land\sqrt{\sinh^{2}z}=\sinh z\lor\sqrt{\cosh z-1}\sqrt{\cosh z+1}=-\sqrt{\cosh^{2}z-1}\land\sqrt{\sinh^{2}z}=-\sinh z\right)\land-\pi<\Im z\leq\pi\\ \Leftrightarrow & \left(\left(0<\Re z\land-\frac{\pi}{2}<\Im z\leq\frac{\pi}{2}\right)\lor\left(\Re z=0\land0\leq\Im z\leq\pi\right)\right)\lor\left(\left(0<\Re z\land\left(-\pi<\Im z\leq-\frac{\pi}{2}\lor\frac{\pi}{2}<\Im z\leq\pi\right)\right)\lor\left(\Re z=0\land\Im z=\pi\right)\right)\\ \Leftrightarrow & \left(0<\Re z\land-\pi<\Im z\leq\pi\right)\lor\left(\Re z=0\land0\leq\Im z\leq\pi\right) \end{align*} のとき恒等写像になる。

(3)

\begin{align*} \tanh^{\bullet}\tanh z & =\tanh^{\bullet}\left(-ii\tanh z\right)\\ & =\tanh^{\bullet}\left(-i\tanh\left(iz\right)\right)\\ & =i\tanh^{\bullet}\left(-\tanh\left(iz\right)\right)\\ & =-i\tanh^{\bullet}\left(\tanh\left(iz\right)\right) \end{align*} これより\(\tanh^{\bullet}\left(\tanh\left(iz\right)\right)=iz\)となるとき、恒等写像となるので、\(-\frac{\pi}{2}<\Re\left(iz\right)<\frac{\pi}{2}\lor\left(\Re\left(iz\right)=-\frac{\pi}{2}\land\Im\left(iz\right)<0\right)\lor\left(\Re\left(iz\right)=\frac{\pi}{2}\land\Im\left(iz\right)>0\right)\)すなわち、\(-\frac{\pi}{2}<\Im z<\frac{\pi}{2}\lor\left(\Im z=\frac{\pi}{2}\land\Re z<0\right)\lor\left(\Im z=-\frac{\pi}{2}\land\Re z>0\right)\)となる。

(4)

\begin{align*} \sinh^{-1,\bullet}\sinh^{-1}z & =\sinh^{\bullet}\frac{1}{\sinh^{-1}z}\\ & =\sinh^{\bullet}\sinh z \end{align*} となるので、\(\sinh^{\bullet}\sinh z\)が恒等写像になる領域と同じになるが、\(\sinh^{-1}z\)が定義できない領域を除くと、
\begin{align*} & z\ne0\land\left(-\frac{\pi}{2}<\Im z<\frac{\pi}{2}\lor\left(\Im z=-\frac{\pi}{2}\land\Re z\leq0\right)\lor\left(\Im z=\frac{\pi}{2}\land0\leq\Re z\right)\right)\\ \Leftrightarrow & \left(z\ne0\land-\frac{\pi}{2}<\Im z<\frac{\pi}{2}\right)\lor\left(\Im z=-\frac{\pi}{2}\land\Re z\leq0\right)\lor\left(\Im z=\frac{\pi}{2}\land0\leq\Re z\right) \end{align*} となる。

(5)

\begin{align*} \cosh^{-1,\bullet}\cosh^{-1}z & =\cosh^{\bullet}\frac{1}{\cosh^{-1}z}\\ & =\cosh^{\bullet}\cosh z \end{align*} となるので、\(\cosh^{\bullet}\cosh z\)が恒等写像になる領域と同じになるが、\(\cosh^{-1}z\)が定義できない領域を除くと、
\begin{align*} & z\ne\frac{\pi}{2}\land\left(\left(0<\Re z\land-\pi<\Im z\leq\pi\right)\lor\left(\Re z=0\land0\leq\Im z\leq\pi\right)\right)\\ \Leftrightarrow & \left(z\ne\frac{\pi}{2}\land0<\Re z\land-\pi<\Im z\leq\pi\right)\lor\left(\Re z=0\land0\leq\Im z\leq\pi\right) \end{align*} となる。

(6)

\begin{align*} \tanh^{-1,\bullet}\tanh^{-1}z & =\tanh^{\bullet}\frac{1}{\tanh^{-1}z}\\ & =\tanh^{\bullet}\tanh z \end{align*} となるので、\(\tanh^{\bullet}\tanh z\)が恒等写像になる領域と同じになるが、\(\tanh^{-1}z\)が定義できない領域を除くと、
\begin{align*} & z\ne0\land\left(-\frac{\pi}{2}<\Im z<\frac{\pi}{2}\lor\left(\Im z=\frac{\pi}{2}\land\Re z<0\right)\lor\left(\Im z=-\frac{\pi}{2}\land\Re z>0\right)\right)\\ \Leftrightarrow & \left(z\ne0\land-\frac{\pi}{2}<\Im z<\frac{\pi}{2}\right)\lor\left(\Im z=\frac{\pi}{2}\land\Re z<0\right)\lor\left(\Im z=-\frac{\pi}{2}\land\Re z>0\right) \end{align*} となる。