求解 x^2+y^2+x=25

$${ x }^{ 2 } + { y }^{ 2+x } = 25$$

求解 y 的值 (复数求解)

$$y=e^{\frac{Im(x)arg(25-x^{2})+iRe(x)arg(25-x^{2})+2iarg(25-x^{2})}{\left(Re(x)\right)^{2}+\left(Im(x)\right)^{2}+4Re(x)+4}-\frac{2i\pi n_{1}Re(x)}{\left(Re(x)\right)^{2}+\left(Im(x)\right)^{2}+4Re(x)+4}-\frac{2\pi n_{1}Im(x)}{\left(Re(x)\right)^{2}+\left(Im(x)\right)^{2}+4Re(x)+4}-\frac{4i\pi n_{1}}{\left(Re(x)\right)^{2}+\left(Im(x)\right)^{2}+4Re(x)+4}}\left(|5-x||x+5|\right)^{\frac{Re(x)-iIm(x)+2}{\left(Re(x)\right)^{2}+\left(Im(x)\right)^{2}+4Re(x)+4}}$$

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

求解 y 的值

$$\left\{\begin{matrix}y=\left(25-x^{2}\right)^{\frac{1}{x+2}}\text{, }&\left(\left(25-x^{2}\right)^{\frac{1}{x+2}}<0\text{ and }x>-5\text{ and }x\neq -2\text{ and }Denominator(x)\text{bmod}2=1\text{ and }|x|<5\right)\text{ or }x=5\text{ or }\left(\left(25-x^{2}\right)^{\frac{1}{x+2}}>0\text{ and }x>-5\text{ and }x\neq -2\text{ and }|x|<5\right)\text{ or }\left(\left(25-x^{2}\right)^{\frac{1}{x+2}}<0\text{ and }Denominator(x)\text{bmod}2=1\text{ and }x>-2\text{ and }x\leq 5\right)\text{ or }\left(Numerator(x+2)\text{bmod}2=1\text{ and }Denominator(x)\text{bmod}2=1\text{ and }|x|>5\right)\text{ or }\left(\left(25-x^{2}\right)^{\frac{1}{x+2}}>0\text{ and }x>-2\text{ and }x\leq 5\right)\\y=-\left(25-x^{2}\right)^{\frac{1}{x+2}}\text{, }&\left(x>-5\text{ and }x\neq -2\text{ and }\left(25-x^{2}\right)^{\frac{1}{x+2}}>0\text{ and }Numerator(x+2)\text{bmod}2=0\text{ and }Denominator(x)\text{bmod}2=1\text{ and }|x|<5\right)\text{ or }\left(x>-5\text{ and }x\neq -2\text{ and }\left(25-x^{2}\right)^{\frac{1}{x+2}}<0\text{ and }Numerator(x+2)\text{bmod}2=0\text{ and }|x|<5\right)\text{ or }\left(\left(25-x^{2}\right)^{\frac{1}{x+2}}<0\text{ and }Numerator(x+2)\text{bmod}2=0\text{ and }x>-2\text{ and }x\leq 5\right)\text{ or }\left(x\neq -2\text{ and }Numerator(x+2)\text{bmod}2=1\text{ and }Numerator(x+2)\text{bmod}2=0\text{ and }Denominator(x)\text{bmod}2=1\text{ and }|x|>5\right)\text{ or }\left(\left(25-x^{2}\right)^{\frac{1}{x+2}}>0\text{ and }Numerator(x+2)\text{bmod}2=0\text{ and }Denominator(x)\text{bmod}2=1\text{ and }x>-2\text{ and }x\leq 5\right)\end{matrix}\right.$$