Solve {l}{c_0^n=c_2^n}{a=c_2^n}{text{Solvefor}btext{where}}{b=a}

$$\left. \begin{array} { l } { c_{0} ^ {n} = c_{2} ^ {n} }\\ { a = c_{2} ^ {n} }\\ { \text{Solve for } b \text{ where} } \\ { b = a } \end{array} \right.$$

Solve for c_0, n, c_2, a, b

$$c_{0}=e^{-\frac{2\pi n_{1}\left(Im(n)+iRe(n)\right)}{\left(Re(n)\right)^{2}+\left(Im(n)\right)^{2}}}\text{, }n_{1}\in \mathrm{Z}\text{, }n\in \mathrm{C}\text{, }c_{2}=1\text{, }a=1\text{, }b=1$$

$$c_{0}=e^{\frac{Im(n)arg(c_{2}^{n})+iRe(n)arg(c_{2}^{n})}{\left(Re(n)\right)^{2}+\left(Im(n)\right)^{2}}-\frac{2\pi n_{1}iRe(n)}{\left(Re(n)\right)^{2}+\left(Im(n)\right)^{2}}-\frac{2\pi n_{1}Im(n)}{\left(Re(n)\right)^{2}+\left(Im(n)\right)^{2}}}\left(|c_{2}^{n}|\right)^{\frac{Re(n)-iIm(n)}{\left(Re(n)\right)^{2}+\left(Im(n)\right)^{2}}}\text{, }n_{1}\in \mathrm{Z}\text{, }n=\frac{i\times 2\pi n_{2}}{\ln(c_{2})}+\log_{c_{2}}\left(b\right)\text{, }n_{2}\in \mathrm{Z}\text{, }c_{2}\in \mathrm{C}\setminus 1,0\text{, }a=b\text{, }b\neq 0$$