Eufisky - The lost book

## i的i次方等于多少?

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$，这是我们从初中就开始熟悉的概念。

$i= (-1)^{1/2} = ((-1)(-1)(-1))^{1/2} =(-1)^{1/2}(-1)^{1/2}(-1)^{1/2}=iii=-i$，

$-1=(-1)^1 = ((-1)^2)^{1/2}=1^{1/2}=1$。

$i^i= \exp(i Ln i) = \exp(i (i\pi/2 + i2k\pi)) = \exp(-\pi/2 - 2k\pi)=\exp(-\pi/2) \exp(-2k\pi)$

$(-2)^{1/3}, \hspace{5 mm}, (-1)^i$

（把你们的解答放在评论里，我的答案后天附在下面）

【答案】

$(-2)^{1/3} = \exp(Ln(-2)/3) = \{-\sqrt[3]{2}, \;\;\; \sqrt[3]{2}(1/2 -i\sqrt{3}/2), \;\;\; \sqrt[3]{2}(1/2 + i\sqrt{3}/2) \}$

$(-1)^i = \exp(iLn(-1))= e^{-\pi+2k\pi} = 0.043213918 \times 535.4916555^k$

$x_0 = -\sqrt[5]{2}$

$x_1 = -\sqrt[5]{2}((\sqrt{5}-1)/4+ i \sqrt{10+2\sqrt{5}}/4), \;\;\; x_2 =-\sqrt[5]{2}(-(\sqrt{5}+1)/4 + i \sqrt{10-2\sqrt{5}}/4)$

$x_3 = -\sqrt[5]{2}(-(\sqrt{5}+1)/4 - i \sqrt{10-2\sqrt{5}}/4),\;\;\; x_4 = -\sqrt[5]{2}((\sqrt{5}-1)/4- i \sqrt{10+2\sqrt{5}}/4),$

$x_n = -\sqrt[5]{2}(\cos(2n\pi/5) +i\sin(2n\pi/5)), \;\;\; n=0,1,2,3,4$

$\cos(2\pi/5)=(\sqrt{5}-1)/4= 0.309016994...$，

$\sin(2\pi/5)=\sqrt{10+2\sqrt{5}}/4= 0.951056516...$，

$\cos(\pi/5)=(\sqrt{5}+1)/4= 0.809016994…$

$\sin(\pi/5)=\sqrt{10-2\sqrt{5}}/4= 0.587785252...$