Solve z=(operatorname{arctg}x)^ln(y-2)

$$z = ( \operatorname { arctg } x ) ^ { \ln ( y - 2 ) }$$

Solve for x (complex solution)

$$x=e^{\frac{arg(z)Im(\ln(y-2))+iarg(z)Re(\ln(y-2))}{\left(Re(\ln(y-2))\right)^{2}+\left(Im(\ln(y-2))\right)^{2}}-\frac{2\pi n_{1}iRe(\ln(y-2))}{\left(Re(\ln(y-2))\right)^{2}+\left(Im(\ln(y-2))\right)^{2}}-\frac{2\pi n_{1}Im(\ln(y-2))}{\left(Re(\ln(y-2))\right)^{2}+\left(Im(\ln(y-2))\right)^{2}}}\left(|z|\right)^{\frac{Re(\ln(y-2))-iIm(\ln(y-2))}{\left(Re(\ln(y-2))\right)^{2}+\left(Im(\ln(y-2))\right)^{2}}}$$

$$n_{1}\in \mathrm{Z}$$

$$y\neq 2$$

Solve for y

$$\left\{\begin{matrix}y=z^{\frac{1}{\ln(\arctan(x))}}+2\text{, }&z^{\frac{1}{\ln(\arctan(x))}}+2>2\text{ and }\arctan(x)>0\text{ and }x\neq \tan(1)\text{ and }z>0\\y\in \mathrm{R}\text{, }&x=-\tan(1)\text{ and }z=-1\text{ and }Denominator(\ln(y-2))\text{bmod}2=1\text{ and }Numerator(\ln(y-2))\text{bmod}2=1\text{ and }y>2\\y>3\text{, }&x=0\text{ and }z=0\\y>2\text{, }&z=1\text{ and }x=\tan(1)\end{matrix}\right.$$