(1)

$$\begin{array}{rl}w& =\sin z\\ & =\frac{e^{iz}-e^{-iz}}{2i}\end{array}$$ より、
$${\left(e^{iz}\right)}^2-2iw\left(e^{iz}\right)-1=0$$ $e^{iz}$について解を1つ選ぶと、
$$e^{iz}=iw+\sqrt{1-w^2}$$ となるので、
$$\begin{array}{rl}{\mathrm{Sin}}^{\bullet }w& =z\\ & =-i\mathrm{Log}(iw+\sqrt{1-w^2})\end{array}$$

(1)補足

$$\sqrt{1-z^2}=\sqrt{1-z}\sqrt{1-z}$$ なので、
$${\mathrm{Sin}}^{\bullet }z=-i\mathrm{Log}(iz+\sqrt{1-z}\sqrt{1-z})$$ としてもいい。

(2)

複素数の範囲でも${\mathrm{Cos}}^{\bullet }z+{\mathrm{Sin}}^{\bullet }z=\frac{\pi }{2}$が成り立つようにする。
$$\begin{array}{rl}{\mathrm{Cos}}^{\bullet }z& =\frac{\pi }{2}-{\mathrm{Sin}}^{\bullet }z\\ & =\frac{\pi }{2}+i\mathrm{Log}(iz+\sqrt{1-z^2})\\ & =-i\mathrm{Log}\left(e^{i\frac{\pi }{2}}\right)+i\mathrm{Log}(iz+\sqrt{1-z^2})\\ & =-i(\mathrm{Log}\left(i\right)-\mathrm{Log}(iz+\sqrt{1-z^2}))\\ & =-i\mathrm{Log}\left(\frac{i}{iz+\sqrt{1-z^2}}\right)(\ast )\\ & =-i\mathrm{Log}\left(\frac{i(-iz+\sqrt{1-z^2})}{(iz+\sqrt{1-z^2})(-iz+\sqrt{1-z^2})}\right)\\ & =-i\mathrm{Log}(z+i\sqrt{1-z^2})\end{array}$$

(2)-2

$$\begin{array}{rl}w& =\cos z\\ & =\frac{e^{iz}+e^{-iz}}{2}\end{array}$$ より、
$${\left(e^{iz}\right)}^2-2w\left(e^{iz}\right)+1=0$$ $e^{iz}$について解を1つ選ぶと、
$$e^{iz}=w+i\sqrt{1-w^2}$$ となるので、
$$\begin{array}{rl}{\mathrm{Cos}}^{\bullet }w& =z\\ & =-i\mathrm{Log}(w+i\sqrt{1-w^2})\end{array}$$

(3)

$$\begin{array}{rl}w& =\tan z\\ & =-i\frac{e^{iz}-e^{-iz}}{e^{iz}+e^{-iz}}\end{array}$$ より、
$$(w+i){\left(e^{iz}\right)}^2+(w-i)=0$$ $e^{iz}$について解を1つ選ぶと、
$$e^{iz}=\frac{\sqrt{1+iw}}{\sqrt{1-iw}}$$ となるので、
$$\begin{array}{rl}{\mathrm{Tan}}^{\bullet }w& =z\\ & =-i\mathrm{Log}\frac{\sqrt{1+iw}}{\sqrt{1-iw}}\\ & =-\frac{i}{2}(\mathrm{Log}(1+iw)-\mathrm{Log}(1-iw))\\ & =\frac{i}{2}(\mathrm{Log}(1-iw)-\mathrm{Log}(1+iw))\end{array}$$

(3)補足

$$\begin{array}{rl}\mathrm{Log}(1-2)-\mathrm{Log}(1+2)& =\mathrm{Log}(-1))-\mathrm{Log}\left(3\right)\\ & =i\pi -\mathrm{Log}\left(3\right)\\ & \ne \mathrm{Log}\frac{1-2}{1+2}\\ & =-\ln 3-i\pi \\ & =\ln \frac{1}{3}+i\mathrm{Arg}(-\frac{1}{3})\\ & =\mathrm{Log}(-\frac{1}{3})\\ & =\mathrm{Log}\frac{1-2}{1+2}\end{array}$$ となるので、
$$\mathrm{Log}(1-iw)-\mathrm{Log}(1+iw)\ne \mathrm{Log}\frac{1-iw}{1+iw}$$ である。

(4)

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

(4)-2

$$\begin{array}{rl}w& ={\sin }^{-1}z\\ & =\frac{2i}{e^{iz}-e^{-iz}}\end{array}$$ より、
$$w{\left(e^{iz}\right)}^2-2i\left(e^{iz}\right)-w=0$$ $e^{iz}$について解を1つ選ぶと、
$$e^{iz}=\frac{i+\sqrt{w^2-1}}{w}$$ となるので、
$$\begin{array}{rl}{\sin }^{-1,\bullet }w& =z\\ & =-i\mathrm{Log}(\frac{i}{w}+\sqrt{1-\frac{1}{w^2}})\end{array}$$

(5)

$$\begin{array}{rl}{\cos }^{-1,\bullet }z& ={\cos }^{-1,\bullet \bullet ,-1,\bullet }\frac{1}{z}\\ & ={\cos }^{\bullet }\frac{1}{z}\\ & =-i\mathrm{Log}(\frac{1}{z}+i\sqrt{1-\frac{1}{z^2}})\end{array}$$

(5)-2

$$\begin{array}{rl}{\cos }^{-1,\bullet }z& =\frac{\pi }{2}-{\sin }^{-1,\bullet }z\\ & =\frac{\pi }{2}+i\mathrm{Log}(\frac{i}{z}+\sqrt{1-\frac{1}{z^2}})\\ & =-i\mathrm{Log}{e}^{i\frac{\pi }{2}}+i\mathrm{Log}(\frac{i}{z}+\sqrt{1-\frac{1}{z^2}})\\ & =-i(\mathrm{Log}i-\mathrm{Log}(\frac{i}{z}+\sqrt{1-\frac{1}{z^2}}))\\ & =-i\left(\mathrm{Log}\frac{i}{\frac{i}{z}+\sqrt{1-\frac{1}{z^2}}}\right)\\ & =-i\left(\mathrm{Log}\frac{i(-\frac{i}{z}+\sqrt{1-\frac{1}{z^2}})}{(\frac{i}{z}+\sqrt{1-\frac{1}{z^2}})(-\frac{i}{z}+\sqrt{1-\frac{1}{z^2}})}\right)\\ & =-i(\mathrm{Log}\frac{1}{z}+i\sqrt{1-\frac{1}{z^2}})\end{array}$$

(5)-3

$$\begin{array}{rl}w& ={\cos }^{-1}z\\ & =\frac{2}{e^{iz}+e^{-iz}}\end{array}$$ より、
$$w{\left(e^{iz}\right)}^2-2\left(e^{iz}\right)+w=0$$ $e^{iz}$について解を1つ選ぶと、
$$e^{iz}=\frac{1+\sqrt{1-w^2}}{w}$$ となるので、
$$\begin{array}{rl}{\cos }^{-1,\bullet }w& =z\\ & =-i\mathrm{Log}(\frac{1}{w}+i\sqrt{1-\frac{1}{w^2}})\end{array}$$

(6)

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

(6)-2

$$\begin{array}{rl}w& ={\tan }^{-1}z\\ & =i\frac{e^{iz}+e^{-iz}}{e^{iz}-e^{-iz}}\end{array}$$ より、
$$(w-i){\left(e^{iz}\right)}^2-(w+i)=0$$ $e^{iz}$について解を1つ選ぶと、
$$\begin{array}{rl}{e}^{iz}& =\frac{\sqrt{\frac{w+i}{w}}}{\sqrt{\frac{w-i}{w}}}\\ & =\frac{\sqrt{1+\frac{i}{w}}}{\sqrt{1-\frac{i}{w}}}\end{array}$$ となるので、
$$\begin{array}{rl}{\tan }^{-1,\bullet }w& =z\\ & =-i\mathrm{Log}\left(\frac{\sqrt{1+\frac{i}{w}}}{\sqrt{1-\frac{i}{w}}}\right)\\ & =-\frac{i}{2}(\mathrm{Log}(1+\frac{i}{w})-\mathrm{Log}(1-\frac{i}{w}))\\ & =\frac{i}{2}(\mathrm{Log}(1-\frac{i}{w})-\mathrm{Log}(1+\frac{i}{w}))\end{array}$$

(1)

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

(1)-2

$$\begin{array}{rl}w& =\sinh z\\ & =\frac{e^z-e^{-z}}{2}\end{array}$$ より、
$${\left(e^z\right)}^2-2w\left(e^z\right)-1=0$$ $e^z$について解を1つ選ぶと、
$$e^z=w+\sqrt{w^2+1}$$ となるので、
$$\begin{array}{rl}{\mathrm{Sinh}}^{\bullet }w& =z\\ & =\mathrm{Log}(w+\sqrt{w^2+1})\end{array}$$

(1)補足

$$\sqrt{z^2+1}=\sqrt{z-i}\sqrt{z+i}$$ なので、
$${\mathrm{Sinh}}^{\bullet }z=\mathrm{Log}(z+\sqrt{z-i}\sqrt{z+i})$$ としてもいい。

(2)

$$\begin{array}{rl}w& =\cosh z\\ & =\frac{e^z+e^{-z}}{2}\end{array}$$ より、
$${\left(e^z\right)}^2-2w\left(e^z\right)+1=0$$ $e^{iz}$について解を1つ選ぶと、
$$e^z=w+\sqrt{w-1}\sqrt{w+1}$$ となるので、
$$\begin{array}{rl}{\mathrm{Cosh}}^{\bullet }w& =z\\ & =\mathrm{Log}(w+\sqrt{w-1}\sqrt{w+1})\end{array}$$

(2)補足

$$\sqrt{z-1}\sqrt{z+1}\ne \sqrt{z^2-1}$$ なので、分岐截線の都合上、
$${\mathrm{Cosh}}^{\bullet }z=\mathrm{Log}(z+\sqrt{z^2-1})$$ としない。

(3)

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

(3)-2

$$\begin{array}{rl}w& =\tanh z\\ & =\frac{e^z-e^{-z}}{e^z+e^{-z}}\end{array}$$ より、
$$(w-1){\left(e^z\right)}^2+(w+1)=0$$ $e^z$について解を1つ選ぶと、
$$e^z=\frac{\sqrt{1+w}}{\sqrt{1-w}}$$ となるので、
$$\begin{array}{rl}{\mathrm{Tanh}}^{\bullet }w& =z\\ & =\mathrm{Log}\frac{\sqrt{1+w}}{\sqrt{1-w}}\\ & =\frac{1}{2}(\mathrm{Log}(1+w)-\mathrm{Log}(1-w))\end{array}$$

(4)

$$\begin{array}{rl}{\sinh }^{-1,\bullet }z& ={\sinh }^{-1,\bullet \bullet ,-1,\bullet }\frac{1}{z}\\ & ={\mathrm{Sinh}}^{\bullet }\frac{1}{z}\\ & =\mathrm{Log}(\frac{1}{z}+\sqrt{\frac{1}{z^2}+1})\end{array}$$

(4)-2

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

(4)-3

$$\begin{array}{rl}w& ={\sinh }^{-1}z\\ & =\frac{2}{e^z-e^{-z}}\end{array}$$ より、
$$w{\left(e^z\right)}^2-2\left(e^z\right)-w=0$$ $e^z$について解を1つ選ぶと、
$$e^z=\frac{1+\sqrt{1+w^2}}{w}$$ となるので、
$$\begin{array}{rl}{\sinh }^{-1,\bullet }w& =z\\ & =\mathrm{Log}(\frac{1}{w}+\sqrt{\frac{1}{w^2}+1})\end{array}$$

(5)

$$\begin{array}{rl}{\cosh }^{-1,\bullet }z& ={\cosh }^{-1,\bullet \bullet ,-1,\bullet }\frac{1}{z}\\ & ={\mathrm{Cosh}}^{\bullet }\frac{1}{z}\\ & =\mathrm{Log}(\frac{1}{z}+\sqrt{\frac{1}{z}-1}\sqrt{\frac{1}{z}+1})\end{array}$$

(5)-2

$$\begin{array}{rl}w& ={\cosh }^{-1}z\\ & =\frac{2}{e^z+e^{-z}}\end{array}$$ より、
$$w{\left(e^z\right)}^2-2\left(e^z\right)+w=0$$ $e^z$について解を選ぶと、
$$\begin{array}{rl}{e}^z& =\frac{1}{w}+\sqrt{\frac{1-w}{w}}\sqrt{\frac{1+w}{w}}\\ & =\frac{1}{w}+\sqrt{\frac{1}{w}-1}\sqrt{\frac{1}{w}+1}\end{array}$$ となるので、
$$\begin{array}{rl}{\cosh }^{-1,\bullet }w& =z\\ & =\mathrm{Log}(\frac{1}{w}+\sqrt{\frac{1}{w}-1}\sqrt{\frac{1}{w}+1})\end{array}$$

(6)

$$\begin{array}{rl}{\tanh }^{-1,\bullet }z& ={\tanh }^{-1,\bullet \bullet ,-1,\bullet }\frac{1}{z}\\ & ={\mathrm{Tanh}}^{\bullet }\frac{1}{z}\\ & =\frac{1}{2}(\mathrm{Log}(1+\frac{1}{z})-\mathrm{Log}(1-\frac{1}{z}))\end{array}$$

(6)-2

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

(6)-3

$$\begin{array}{rl}w& ={\tanh }^{-1}z\\ & =\frac{e^z+e^{-z}}{e^z-e^{-z}}\end{array}$$ より、
$$(w-1){\left(e^{iz}\right)}^2-(w+1)=0$$ $e^z$について解を1つ選ぶと、
$$\begin{array}{rl}{e}^{iz}& =\frac{\sqrt{\frac{w+1}{w}}}{\sqrt{\frac{w-1}{w}}}\\ & ==\frac{\sqrt{1+\frac{1}{w}}}{\sqrt{1-\frac{1}{w}}}\end{array}$$ となるので、
$$\begin{array}{rl}{\tanh }^{-1,\bullet }w& =z\\ & =\mathrm{Log}\left(\frac{\sqrt{1+\frac{1}{w}}}{\sqrt{1-\frac{1}{w}}}\right)\\ & =\frac{1}{2}(\mathrm{Log}(1+\frac{1}{w})-\mathrm{Log}(1-\frac{1}{w}))\end{array}$$