(1)

$$\begin{array}{rl}{\mathrm{Sin}}^{\bullet }\left(iz\right)& ={\mathrm{Sin}}^{\bullet }(i\sinh {\mathrm{Sinh}}^{\bullet }z)\\ & ={\mathrm{Sin}}^{\bullet }\sin \left(i{\mathrm{Sinh}}^{\bullet }z\right)\\ & =i{\mathrm{Sinh}}^{\bullet }z\end{array}$$ $i{\mathrm{Sinh}}^{\bullet }z$の値域は${\mathrm{Sin}}^{\bullet }\sin$が恒等写像になるので成り立つ。

(1)-2

$$\begin{array}{rl}{\mathrm{Sin}}^{\bullet }\left(iz\right)& =-i\mathrm{Log}(i\left(iz\right)+\sqrt{1-{\left(iz\right)}^2})\\ & =-i\mathrm{Log}(-z+\sqrt{1+z^2})\\ & =-i{\mathrm{Sinh}}^{\bullet }(-z)\\ & =i{\mathrm{Sinh}}^{\bullet }z\end{array}$$

(2)

$$\begin{array}{rl}{\mathrm{Cos}}^{\bullet }z& ={\mathrm{Cos}}^{\bullet }\cosh {\mathrm{Cosh}}^{\bullet }z\\ & ={\mathrm{Cos}}^{\bullet }\cos \left(i{\mathrm{Cosh}}^{\bullet }z\right)\\ & ={\mathrm{Cos}}^{\bullet }\cos (-i{\mathrm{Cosh}}^{\bullet }z)\\ & \ne -i{\mathrm{Cosh}}^{\bullet }z\end{array}$$ $-i{\mathrm{Cosh}}^{\bullet }z$の値域は${\mathrm{Cos}}^{\bullet }\cos$が恒等写像になっていない。

(2)-2

$$\begin{array}{rl}{\mathrm{Cos}}^{\bullet }z& =-i\mathrm{Log}(z+i\sqrt{1-z^2})\\ & =-i\mathrm{Log}(z+i\sqrt{1-z}\sqrt{1+z})\\ & \ne -i\mathrm{Log}(z+\sqrt{z-1}\sqrt{z+1})\\ & =-i{\mathrm{Cosh}}^{\bullet }z\end{array}$$

(2)-3 反例

$$\begin{array}{rl}{\mathrm{Cos}}^{\bullet }(-1)& =\pi \\ & =-i{\mathrm{Cosh}}^{\bullet }(-1)\end{array}$$ $$\begin{array}{rl}{\mathrm{Cos}}^{\bullet }\left(0\right)& =\frac{\pi }{2}\\ & =-i{\mathrm{Cosh}}^{\bullet }\left(0\right)\end{array}$$ となるが、$\frac{\pi }{2}<\mathrm{Arg}\sqrt{1-z}\vee \mathrm{Arg}\sqrt{1-z}\le -\frac{\pi }{2}$ととると、
$$\begin{array}{rl}{\mathrm{Cos}}^{\bullet }\left(2\right)& =i{\mathrm{Cosh}}^{\bullet }\left(2\right)\\ & \ne -i{\mathrm{Cosh}}^{\bullet }\left(2\right)\end{array}$$ となるので成り立たない。

(3)

$$\begin{array}{rl}{\mathrm{Tan}}^{\bullet }\left(iz\right)& ={\mathrm{Tan}}^{\bullet }(i\tanh {\mathrm{Tanh}}^{\bullet }z)\\ & ={\mathrm{Tan}}^{\bullet }\tan \left(i{\mathrm{Tanh}}^{\bullet }z\right)\\ & =i{\mathrm{Tanh}}^{\bullet }z\end{array}$$ $i{\mathrm{Tanh}}^{\bullet }z$の値域は${\mathrm{Tan}}^{\bullet }\tan$が恒等写像になるので成り立つ。

(3)-2

$$\begin{array}{rl}{\mathrm{Tan}}^{\bullet }\left(iz\right)& =\frac{i}{2}(\mathrm{Log}(1-i\left(iz\right))-\mathrm{Log}(1+i\left(iz\right)))\\ & =\frac{i}{2}(\mathrm{Log}(1+z)-\mathrm{Log}(1-z))\\ & =i{\mathrm{Tanh}}^{\bullet }z\end{array}$$

(4)

$$\begin{array}{rl}{\mathrm{Sin}}^{-1,\bullet }\left(iz\right)& ={\mathrm{Sin}}^{-1,\bullet }(i{\sinh }^{-1}{\mathrm{Sinh}}^{-1,\bullet }z)\\ & ={\mathrm{Sin}}^{-1,\bullet }(-{\sin }^{-1}\left(i{\mathrm{Sinh}}^{-1,\bullet }z\right))\\ & ={\mathrm{Sin}}^{-1,\bullet }{\sin }^{-1}(-i{\mathrm{Sinh}}^{-1,\bullet }z)\\ & =-i{\mathrm{Sinh}}^{-1,\bullet }z\end{array}$$ $-i{\sinh }^{-1,\bullet }z$の値域は${\sin }^{-1,\bullet }{\sin }^{-1}$が恒等写像になるので成り立つ。

(4)-2

$$\begin{array}{rl}{\mathrm{Sin}}^{-1,\bullet }\left(iz\right)& ={\mathrm{Sin}}^{\bullet }(-\frac{i}{z})\\ & =-i{\mathrm{Sinh}}^{\bullet }\left(\frac{1}{z}\right)\\ & =-i{\mathrm{Sinh}}^{-1,\bullet }z\end{array}$$

(5)

$$\begin{array}{rl}{\mathrm{Cos}}^{-1,\bullet }z& ={\mathrm{Cos}}^{-1,\bullet }{\cosh }^{-1}{\mathrm{Cosh}}^{-1,\bullet }z\\ & ={\mathrm{Cos}}^{-1,\bullet }{\cos }^{-1}\left(i{\mathrm{Cosh}}^{-1,\bullet }z\right)\\ & ={\mathrm{Cos}}^{-1,\bullet }{\cos }^{-1}(-i{\mathrm{Cosh}}^{-1,\bullet }z)\\ & \ne -i{\mathrm{Cosh}}^{-1,\bullet }z\end{array}$$ $-i{\mathrm{Cosh}}^{-1,\bullet }z$の値域は${\mathrm{Cos}}^{-1,\bullet }{\cos }^{-1}$が恒等写像になっていない。

(6)

$$\begin{array}{rl}{\mathrm{Tan}}^{-1,\bullet }\left(iz\right)& ={\mathrm{Tan}}^{-1,\bullet }(i{\tanh }^{-1}{\mathrm{Tanh}}^{-1,\bullet }z)\\ & ={\mathrm{Tan}}^{-1,\bullet }(-{\tan }^{-1}\left(i{\mathrm{Tanh}}^{-1,\bullet }z\right))\\ & ={\mathrm{Tan}}^{-1,\bullet }{\tan }^{-1}(-i{\mathrm{Tanh}}^{-1,\bullet }z)\\ & =-i{\mathrm{Tanh}}^{-1,\bullet }z\end{array}$$ $-i{\mathrm{Tanh}}^{-1,\bullet }z$の値域は${\mathrm{Tan}}^{-1,\bullet }{\tan }^{-1}$が恒等写像になるので成り立つ。

(6)-2

$$\begin{array}{rl}{\mathrm{Tan}}^{-1,\bullet }\left(iz\right)& ={\mathrm{Tan}}^{\bullet }(-\frac{i}{z})\\ & =-i{\mathrm{Tanh}}^{\bullet }\left(\frac{1}{z}\right)\\ & =-i{\mathrm{Tanh}}^{-1,\bullet }z\end{array}$$

(1)

$$\begin{array}{rl}{\mathrm{Sinh}}^{\bullet }\left(iz\right)& ={\mathrm{Sinh}}^{\bullet }(i\sin {\mathrm{Sin}}^{\bullet }z)\\ & ={\mathrm{Sinh}}^{\bullet }\sinh \left(i{\mathrm{Sin}}^{\bullet }z\right)\\ & =i{\mathrm{Sin}}^{\bullet }\left(z\right)\end{array}$$ $i{\mathrm{Sin}}^{\bullet }z$の値域は${\mathrm{Sinh}}^{\bullet }\sinh$が恒等写像になるので成り立つ。

(1)-2

$$\begin{array}{rl}{\mathrm{Sinh}}^{\bullet }\left(iz\right)& =\mathrm{Log}(iz+\sqrt{1+{\left(iz\right)}^2})\\ & =i(-i)\mathrm{Log}(iz+\sqrt{1-z^2})\\ & =i{\mathrm{Sin}}^{\bullet }z\end{array}$$

(2)

$$\begin{array}{rl}{\mathrm{Cosh}}^{\bullet }z& ={\mathrm{Cosh}}^{\bullet }\cos {\mathrm{Cos}}^{\bullet }z\\ & ={\mathrm{Cosh}}^{\bullet }\cosh \left(i{\mathrm{Cos}}^{\bullet }z\right)\\ & \ne i{\mathrm{Cos}}^{\bullet }z\end{array}$$ $i{\mathrm{Cos}}^{\bullet }z$の値域は${\mathrm{Cosh}}^{\bullet }\cosh$が恒等写像になっていない。

(2)-2

$$\begin{array}{rl}{\mathrm{Cosh}}^{\bullet }z& =\mathrm{Log}(z+\sqrt{z-1}\sqrt{z+1})\\ & \ne \mathrm{Log}(z+i\sqrt{1-z}\sqrt{1+z})\\ & =\mathrm{Log}(z+i\sqrt{1-z^2})\\ & =i(-i)\mathrm{Log}(z+i\sqrt{1-z^2})\\ & =i{\mathrm{Cos}}^{\bullet }z\end{array}$$

(3)

$$\begin{array}{rl}{\mathrm{Tanh}}^{\bullet }\left(iz\right)& ={\mathrm{Tanh}}^{\bullet }(i\tan {\mathrm{Tan}}^{\bullet }z)\\ & ={\mathrm{Tanh}}^{\bullet }\tanh \left(i{\mathrm{Tan}}^{\bullet }z\right)\\ & =i{\mathrm{Tan}}^{\bullet }z\end{array}$$ $i{\mathrm{Tan}}^{\bullet }z$の値域は${\mathrm{Tanh}}^{\bullet }\tanh$が恒等写像になるので成り立つ。

(3)-2

$$\begin{array}{rl}{\mathrm{Tanh}}^{\bullet }\left(iz\right)& =\frac{1}{2}(\mathrm{Log}(1+iz)-\mathrm{Log}(1-iz))\\ & =i\frac{i}{2}(\mathrm{Log}(1-iz)-\mathrm{Log}(1+iz))\\ & =i{\mathrm{Tanh}}^{\bullet }z\end{array}$$

(4)

$$\begin{array}{rl}{\mathrm{Sinh}}^{-1,\bullet }\left(iz\right)& ={\mathrm{Sinh}}^{-1,\bullet }(i{\sin }^{-1}{\mathrm{Sin}}^{-1,\bullet }z)\\ & ={\mathrm{Sinh}}^{-1,\bullet }(-{\sinh }^{-1}\left(i{\mathrm{Sin}}^{-1,\bullet }z\right))\\ & ={\mathrm{Sinh}}^{-1,\bullet }{\sinh }^{-1}(-i{\mathrm{Sin}}^{-1,\bullet }z)\\ & =-i{\mathrm{Sin}}^{-1,\bullet }z\end{array}$$ $-i{\mathrm{Sin}}^{-1,\bullet }z$の値域は${\mathrm{Sinh}}^{-1,\bullet }{\sinh }^{-1}$が恒等写像になるので成り立つ。

(4)-2

$$\begin{array}{rl}{\mathrm{Sinh}}^{-1,\bullet }\left(iz\right)& ={\mathrm{Sinh}}^{\bullet }(-\frac{i}{z})\\ & =-i{\mathrm{Sin}}^{\bullet }\left(\frac{1}{z}\right)\\ & =-i{\mathrm{Sin}}^{-1,\bullet }\left(z\right)\end{array}$$

(5)

$$\begin{array}{rl}{\mathrm{Cosh}}^{-1,\bullet }z& ={\mathrm{Cosh}}^{-1,\bullet }{\cos }^{-1}{\mathrm{Cos}}^{-1,\bullet }z\\ & ={\mathrm{Cosh}}^{-1,\bullet }{\cosh }^{-1}\left(i{\mathrm{Cos}}^{-1,\bullet }z\right)\\ & \ne i{\mathrm{Cos}}^{-1,\bullet }z\end{array}$$ $i{\mathrm{Cos}}^{-1,\bullet }z$の値域は${\mathrm{Cosh}}^{-1,\bullet }{\cosh }^{-1}$が恒等写像になっていない。

(5)-2

$$\begin{array}{rl}{\mathrm{Cosh}}^{-1,\bullet }\left(z\right)& ={\mathrm{Cosh}}^{\bullet }\left(\frac{1}{z}\right)\\ & \ne i{\mathrm{Cos}}^{\bullet }\left(\frac{1}{z}\right)\\ & =i{\mathrm{Cos}}^{-1,\bullet }\left(z\right)\end{array}$$

(6)

$$\begin{array}{rl}{\mathrm{Tanh}}^{-1,\bullet }\left(iz\right)& ={\mathrm{Tanh}}^{-1,\bullet }(i{\tan }^{-1}{\mathrm{Tan}}^{-1,\bullet }z)\\ & ={\mathrm{Tanh}}^{-1,\bullet }(-{\tanh }^{-1}\left(i{\mathrm{Tan}}^{-1,\bullet }z\right))\\ & ={\mathrm{Tanh}}^{-1,\bullet }{\tanh }^{-1}(-i{\mathrm{Tan}}^{-1,\bullet }z)\\ & =-i{\mathrm{Tan}}^{-1,\bullet }z\end{array}$$ $-i{\mathrm{Tan}}^{-1,\bullet }z$の値域は${\mathrm{Tanh}}^{-1,\bullet }{\tanh }^{-1}$が恒等写像になるので成り立つ。

(6)-2

$$\begin{array}{rl}{\mathrm{Tanh}}^{-1,\bullet }\left(iz\right)& ={\mathrm{Tanh}}^{\bullet }(-\frac{i}{z})\\ & =-i{\mathrm{Tan}}^{\bullet }\left(\frac{1}{z}\right)\\ & =-i{\mathrm{Tan}}^{-1,\bullet }\left(z\right)\end{array}$$