求解 sqrt[x]{z^2}=x

$$\sqrt[ x ]{ { z }^{ 2 } } =x$$

求解 z 的值 (复数求解)

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

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

$$x\neq 0$$

求解 z 的值

$$\left\{\begin{matrix}z=\sqrt{x^{x}}\text{; }z=-\sqrt{x^{x}}\text{, }&x\neq 0\text{ and }\left(Denominator(x)\text{bmod}2=1\text{ or }x>0\right)\text{ and }x^{x}\geq 0\text{ and }\left(Numerator(x)\text{bmod}2=1\text{ or }x>0\right)\\z=\sqrt{-x^{x}}\text{; }z=-\sqrt{-x^{x}}\text{, }&\left(x>0\text{ and }x^{x}<0\text{ and }Denominator(x)\text{bmod}2=0\right)\text{ or }\left(x<0\text{ and }Denominator(x)\text{bmod}2=1\text{ and }x^{x}\leq 0\text{ and }Denominator(x)\text{bmod}2=0\text{ and }Numerator(x)\text{bmod}2=1\right)\end{matrix}\right.$$