# 訊息

## English

(1)
\begin{eqnarray} i^0 &=& 1\\ i^1 &=& i \\ i^2 &=& -1 \\ i^3 &=& -i \\ i^4 &=& 1 \\ i^5 &=& i \\ i^6 &=& -1 \\ i^7 &=& -i \\ \end{eqnarray}

(2)
\begin{align} i^{1/2} = cos(\pi/4) + i*sin(\pi/4) = \frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}} \end{align}

(3)
\begin{eqnarray} 左邊 && (i^{1/2} )^2 = i^{1/2 * 2} = i \\ 右邊 && (\frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}} )^2 = 1/2 + 2*\frac{1}{2} i - 1/2 = i \\ \end{eqnarray}

(4)
\begin{eqnarray} e^{i x} = cos(x) + i*sin(x) \end{eqnarray}

(5)
\begin{eqnarray} i^x &=& e^{x * ln(i) } \\ &=& e^{x*(ln|i| + i * arg (i))} \\ &=& e^{x*(ln(1) + i * (\pi/2 + 2k \pi)) } \\ &=& e^{x*ln(1)}*e^{i * x * (\pi/2 + 2k \pi)} \\ &=& 1 * e^{i * x * (\pi/2 + 2k \pi)} \\ &=& e^{i x (\pi/2 + 2k \pi)} \end{eqnarray}

# 尤拉定理

(6)
\begin{eqnarray} e^{i \theta} = cos(\theta) + i*sin(\theta) \end{eqnarray}

$\theta = \pi$ 時，可得到下列算式：

(7)
\begin{eqnarray} e^{i \pi} + 1 = 0 \end{eqnarray}

# 參考文獻

1. 棣美弗定理與 Euler 公式, 林琦焜
2. 作者: scorpioeric (常行一式) 看板: Math, 標題: Re: [問題] i的i次方?, 時間: Wed Aug 2 00:23:49 2006