다음을 풀어보세요: a^n=z

$$a ^ { n } = z$$

a에 대한 해 (complex solution)

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

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

n에 대한 해 (complex solution)

$$\left\{\begin{matrix}n=\frac{2\pi n_{1}i}{\ln(a)}+\log_{a}\left(z\right)\text{, }n_{1}\in \mathrm{Z}\text{, }&z\neq 0\text{ and }a\neq 1\text{ and }a\neq 0\\n\in \mathrm{C}\text{, }&\left(a=0\text{ and }z=0\right)\text{ or }\left(a=1\text{ and }z=1\right)\end{matrix}\right.$$

a에 대한 해

$$\left\{\begin{matrix}a=z^{\frac{1}{n}}\text{, }&\left(Numerator(n)\text{bmod}2=1\text{ and }Denominator(n)\text{bmod}2=1\text{ and }z<0\text{ and }z^{\frac{1}{n}}\neq 0\right)\text{ or }\left(z=0\text{ and }n>0\right)\text{ or }\left(z>0\text{ and }n\neq 0\right)\\a=-z^{\frac{1}{n}}\text{, }&\left(z<0\text{ and }Numerator(n)\text{bmod}2=1\text{ and }Numerator(n)\text{bmod}2=0\text{ and }Denominator(n)\text{bmod}2=1\text{ and }z^{\frac{1}{n}}\neq 0\right)\text{ or }\left(n\neq 0\text{ and }z>0\text{ and }Numerator(n)\text{bmod}2=0\text{ and }Denominator(n)\text{bmod}2=1\right)\text{ or }\left(Numerator(n)\text{bmod}2=0\text{ and }z=0\text{ and }n>0\right)\text{ or }\left(z>0\text{ and }n\neq 0\text{ and }z^{\frac{1}{n}}<0\text{ and }Numerator(n)\text{bmod}2=0\right)\\a\neq 0\text{, }&n=0\text{ and }z=1\end{matrix}\right.$$

n에 대한 해

$$\left\{\begin{matrix}n=\log_{a}\left(z\right)\text{, }&z>0\text{ and }a\neq 1\text{ and }a>0\\n\in \mathrm{R}\text{, }&\left(a=1\text{ and }z=1\right)\text{ or }\left(a=-1\text{ and }z=-1\text{ and }Denominator(n)\text{bmod}2=1\text{ and }Numerator(n)\text{bmod}2=1\right)\\n>0\text{, }&a=0\text{ and }z=0\end{matrix}\right.$$