求解 E=mc^ty^2

$$E = m c ^ { t y ^ { 2 } }$$

求解 c 的值 (复数求解)

$$\left\{\begin{matrix}c=\left(\frac{|E|}{|m|}\right)^{\frac{Re(t)Re(y^{2})-Im(t)Im(y^{2})-iRe(t)Im(y^{2})-iIm(t)Re(y^{2})}{\left(\left(Re(t)\right)^{2}+\left(Im(t)\right)^{2}\right)\left(\left(Re(y)\right)^{2}+\left(Im(y)\right)^{2}\right)^{2}}}e^{\frac{2\pi n_{1}iIm(t)Im(y^{2})}{\left(\left(Re(t)\right)^{2}+\left(Im(t)\right)^{2}\right)\left(\left(Re(y)\right)^{2}+\left(Im(y)\right)^{2}\right)^{2}}-\frac{2\pi n_{1}iRe(t)Re(y^{2})}{\left(\left(Re(t)\right)^{2}+\left(Im(t)\right)^{2}\right)\left(\left(Re(y)\right)^{2}+\left(Im(y)\right)^{2}\right)^{2}}-\frac{2\pi n_{1}Re(t)Im(y^{2})}{\left(\left(Re(t)\right)^{2}+\left(Im(t)\right)^{2}\right)\left(\left(Re(y)\right)^{2}+\left(Im(y)\right)^{2}\right)^{2}}-\frac{2\pi n_{1}Im(t)Re(y^{2})}{\left(\left(Re(t)\right)^{2}+\left(Im(t)\right)^{2}\right)\left(\left(Re(y)\right)^{2}+\left(Im(y)\right)^{2}\right)^{2}}+\frac{arg(\frac{E}{m})Re(t)Im(y^{2})+arg(\frac{E}{m})Im(t)Re(y^{2})+iarg(\frac{E}{m})Re(t)Re(y^{2})-iarg(\frac{E}{m})Im(t)Im(y^{2})}{\left(\left(Re(t)\right)^{2}+\left(Im(t)\right)^{2}\right)\left(\left(Re(y)\right)^{2}+\left(Im(y)\right)^{2}\right)^{2}}}\text{, }n_{1}\in \mathrm{Z}\text{, }&m\neq 0\\c\in \mathrm{C}\text{, }&E=0\text{ and }m=0\end{matrix}\right.$$

求解 E 的值

$$E=mc^{ty^{2}}$$

$$c>0\text{ or }\left(c=0\text{ and }t>0\text{ and }y\neq 0\right)\text{ or }\left(c<0\text{ and }Denominator(ty^{2})\text{bmod}2=1\right)$$