複素数のn乗を求める(ド・モアブルの定理)

ド・モアブルの定理
n が整数のとき

(\cos\theta + i \sin\theta)^n = \cos n \theta + i \sin n \theta
\begin{align*}
& \textcolor{orange}{\small\phase{ ド・モアブルの定理!}}\\
&\textcolor{white}{=} \left(\cos\dfrac{\pi}{3}+i\sin\dfrac{\pi}{3}\right)^{\textcolor{red}{4}}\\\\
& \textcolor{orange}{\small\phase{ \theta に 4 をかける}}\\
&= \cos \textcolor{red}{4 \times} \dfrac{\pi}{3}+i \sin \textcolor{red}{4 \times} \dfrac{\pi}{3}\\\\
&= \cos\dfrac{4}{3}\pi +i \sin\dfrac{4}{3}\pi\\\\
&= \textcolor{red}{\underline{\underline{\textcolor{black}{\ -\dfrac12 - \dfrac{\sqrt{3}}{2}i\ }}}}

\end{align*}
\begin{align*}
& \textcolor{orange}{\small\phase{ ド・モアブルの定理!}}\\
&\textcolor{white}{=} \left(\cos\dfrac{\pi}{6}+i\sin\dfrac{\pi}{6}\right)^{\textcolor{red}{-6}}\\\\
& \textcolor{orange}{\small\phase{ \theta に -6 をかける}}\\
&= \cos \textcolor{red}{(-6) \times} \dfrac{\pi}{6}+i \sin \textcolor{red}{(-6) \times} \dfrac{\pi}{6}\\\\
&= \cos (-\pi) +i \sin (-\pi)\\\\
& \textcolor{orange}{\small\phase{ \cos(-\theta)=\cos \theta}} \textcolor{orange}{\small\phase{ \sin(-\theta)=-\sin \theta}}\\
&= \cos \pi -i \sin \pi\\\\
&= -1 - i \cdot 0\\
&= \textcolor{red}{\underline{\underline{\textcolor{black}{\ -1\ }}}}
\end{align*}
\begin{align*}
& \textcolor{orange}{\small\phase{ ド・モアブルの定理!}}\\
&\textcolor{white}{=} \left(\cos\dfrac{\pi}{30}+i\sin\dfrac{\pi}{30}\right)^{\textcolor{red}{5}}\\\\
& \textcolor{orange}{\small\phase{ \theta に 5 をかける}}\\
&= \cos \textcolor{red}{5 \times} \dfrac{\pi}{30}+i \sin \textcolor{red}{5 \times} \dfrac{\pi}{30}\\\\
&= \cos \dfrac{\pi}{6} +i \sin \dfrac{\pi}{6}\\\\
&= \textcolor{red}{\underline{\underline{\textcolor{black}{\ \dfrac{\sqrt{3}}{2} + \dfrac12 i\ }}}}
\end{align*}
\begin{align*}
& \textcolor{orange}{\small\phase{ ド・モアブルの定理!}}\\
&\textcolor{white}{=} \left(\cos\dfrac{2}{3}\pi+i\sin\dfrac{2}{3}\pi\right)^{\textcolor{red}{-5}}\\\\
& \textcolor{orange}{\small\phase{ \theta に -5 をかける}}\\
&= \cos \textcolor{red}{(-5) \times} \dfrac{2}{3}\pi+i \sin \textcolor{red}{(-5) \times} \dfrac{2}{3}\pi\\\\
&= \cos \left(-\dfrac{10}{3}\pi\right) +i \sin \left(-\dfrac{10}{3}\pi\right)\\\\
& \textcolor{orange}{\small\phase{ \cos(-\theta)=\cos \theta}} \textcolor{orange}{\small\phase{ \sin(-\theta)=-\sin \theta}}\\
&= \cos \dfrac{10}{3}\pi -i \sin \dfrac{10}{3}\pi\\\\
& \textcolor{orange}{\scriptsize\phase{ 三角関数は\ 2\pi \times n\ 足しても引いても良い!}}\\&= \cos \left(\dfrac{10}{3}\pi-\dfrac{6}{3}\pi\right) -i \sin \left(\dfrac{10}{3}\pi-\dfrac{6}{3}\pi\right)\\\\
&= \cos \dfrac{4}{3}\pi -i \sin \dfrac{4}{3}\pi\\\\
&= \textcolor{red}{\underline{\underline{\textcolor{black}{\ -\dfrac12 +\dfrac{\sqrt{3}}{2} i\ }}}}
\end{align*}

\begin{align*}
& \textcolor{orange}{\small\phase{ ド・モアブルの定理!}}\\
&\textcolor{white}{=} \left\{2\left(\cos\dfrac{\pi}{18}+i\sin\dfrac{\pi}{18}\right)\right\}^{6}\\\\
& \textcolor{orange}{\small\phase{ (ab)^n = a^nb^n}}\\
&= 2^6\left(\cos\dfrac{\pi}{18}+i\sin\dfrac{\pi}{18}\right)^{\textcolor{red}{6}}\\\\
& \textcolor{orange}{\small\phase{ \theta に 6 をかける}}\\
&= 64\left(\cos \textcolor{red}{6 \times} \dfrac{\pi}{18}+i \sin \textcolor{red}{6 \times} \dfrac{\pi}{18}\right)\\\\
&= 64\left(\cos \dfrac{\pi}{3} +i \sin \dfrac{\pi}{3}\right)\\\\
&= 64\left(\dfrac12 +i \cdot \dfrac{\sqrt{3}}{2}\right)\\\\
&= \textcolor{red}{\underline{\underline{\textcolor{black}{\ 32+32\sqrt{3}\,i\ }}}}
\end{align*}
\begin{align*}
& \textcolor{orange}{\small\phase{ 極形式ではない⇒修正}}\\
&\textcolor{white}{=} \left(\cos\dfrac{\pi}{6}\pi-i\sin\dfrac{\pi}{6}\right)^{3}\\\\
& \textcolor{orange}{\small\phase{ \cos(-\theta)=\cos \theta}} \textcolor{orange}{\small\phase{ \sin(-\theta) = -\sin\theta}}\\
&= \left\{\cos\left(-\dfrac{\pi}{6}\right)+i\sin\left(-\dfrac{\pi}{6}\right)\right\}^{\textcolor{red}{3}}\\\\
& \textcolor{orange}{\small\phase{ ド・モアブルの定理!}}\\
& \textcolor{orange}{\small\phase{ \theta に 3 をかける}}\\
&= \cos \textcolor{red}{3 \times} \left(-\dfrac{\pi}{6}\right)+i \sin \textcolor{red}{3 \times} \left(-\dfrac{\pi}{6}\right)\\\\
&= \cos \left(-\dfrac{\pi}{2}\right) +i \sin \left(-\dfrac{\pi}{2}\pi\right)\\\\
& \textcolor{orange}{\small\phase{ \cos(-\theta)=\cos \theta}} \textcolor{orange}{\small\phase{ \sin(-\theta) = -\sin\theta}}\\
&= \cos \dfrac{\pi}{2} -i \sin \dfrac{\pi}{2}\\\\
&= 0-i \cdot 1\\\\
&= \textcolor{red}{\underline{\underline{\textcolor{black}{\ -i\ }}}}
\end{align*}
  • 20210515…初版公開。問題数(6)

コメントを残す

メールアドレスが公開されることはありません。 が付いている欄は必須項目です