46403 再談「圓內接正多邊形的奇偶弦長冪次定和」－－兼談108數學課綱之複數教學

### 1. 森棚教官的數學題

$$e^{i\alpha}=\cos\alpha +i \sin\alpha \hbox{ (其中 \alpha 為實數)},$$

### 2. 一般性問題的描述

\begin{align} \overline{PA_0}^m-\overline{PA_1}^m+\overline{PA_2}^m-\overline{PA_3}^m+\cdots+(-1)^{n-1} \overline{PA_{n-1}}^m=0,\label{1} \end{align}

\begin{align} \sum_{k=0}^{n-1}(-1)^k \overline{PA_k}^m=0.\label{2} \end{align}

### 3. 善用棣美弗定理

$$\cos(\alpha +\beta )+i \sin(\alpha +\beta )=(\cos\alpha +i \sin\alpha) (\cos \beta +i\sin \beta),$$

$$\cos \frac{2\pi k}n+i\sin \frac{2\pi k}n.$$

\begin{align} \overline{PA_k}=2 \sin\Big(\frac{\theta}2+\frac{\pi k}n\Big)；\label{3} \end{align}

$$\overline{PA_k}=2\sin\Big(\pi -\big(\frac \theta 2+\frac{\pi k}n\big)\Big)=2\sin\Big(\frac\theta 2+\frac {\pi k}n\Big),$$

$$z=\cos\Big(\frac\theta 2+\frac{\pi k}n\Big)+i\sin\Big(\frac \theta 2+\frac{\pi k}n\Big),$$

$$z^{-1}=\cos\Big(\frac\theta 2+\frac{\pi k}n\Big)-i\sin\Big(\frac \theta 2+\frac{\pi k}n\Big),$$

$$\overline{PA_k}=2 \sin\Big(\frac \theta 2+\frac{\pi k}2\Big)=i (z^{-1}-z).$$

$$z=\Big(\cos \frac \theta 2+i\sin \frac\theta 2\Big) \Big(\cos \frac \pi n+i\sin \frac \pi n\Big)^k=u v^k,$$

$$u=\cos \frac\theta 2+i\sin \frac\theta 2\quad \hbox{而}\quad v=\cos \frac \pi n+i\sin \frac \pi n.$$

\begin{align} \sum_{k=0}^{n-1}(-1)^k \overline{PA_k}^m=&\sum_{k=0}^{n-1}(-1)^k\Big(i (z^{-1}-z)\Big)^m\nonumber\\ =&\sum_{k=0}^{n-1}\sum_{\ell=0}^m(-1)^k i^m C_\ell^m (z^{-1} )^{m-\ell} (-z)^\ell ,\label{4} \end{align}

\begin{align*} (-1)^k i^m C_\ell^m (z^{-1} )^{m-\ell} (-z)^\ell =\,&i^m C_\ell^m (-1)^\ell (-1)^k z^{2\ell-m}\\ =\,& i^m C_\ell^m (-1)^\ell (-1)^k (u v^k)^{2\ell-m}\\ =\,& i^m C_\ell^m (-1)^\ell u^{2\ell-m} (-v^{2\ell-m})^k, \end{align*}

\begin{align} &\hskip -15pt \sum_{k=0}^{n-1}\sum_{\ell=0}^m(-1)^k i^m C_\ell^m (z^{-1})^{m-\ell} (-z)^\ell\nonumber\\ &=\sum_{\ell=0}^m\Big(i^m C_\ell^m (-1)^\ell u^{2\ell-m} \sum_{k=0}^{n-1}(-v^{2\ell-m} )^k \Big).\label{5} \end{align}

\begin{align*} \sum_{k=0}^{n-1}(-v^{2\ell-m})^k =\frac{1-(-v^{2\ell-m} )^n}{1+v^{2\ell-m}}=\frac{1-(-1)^n (v^n )^{2\ell-m}}{1+v^{2\ell-m}}, \end{align*}

\begin{align} \sum_{k=0}^{n-1}\big(-v^{2\ell-m} \big)^k =\frac{1-(-1)^n (-1)^{2\ell-m}}{1+v^{2\ell-m}}=\frac{1-(-1)^{n+2\ell-m}}{1+v^{2\ell-m}}=0,\label{6} \end{align}