●(準備)加法定理
\[
\begin{align}
\sin(\alpha\pm\beta) &= \sin(\alpha)\cos(\beta)\pm\cos(\alpha)\sin(\beta) \\
\\
\cos(\alpha\pm\beta) &= \cos(\alpha)\cos(\beta)\mp\sin(\alpha)\sin(\beta)
\end{align}
\]
●\(z_1=r_1(\cos(\theta_1)+i\sin(\theta_1)), z_2=r_2(\cos(\theta_2)+i\sin(\theta_2))\) のとき
\[
z_1z_2 = r_1r_2\{\cos(\theta_1+\theta_2)+i\sin(\theta_1+\theta_2)\} \tag{12}
\]
を証明せよ。
(証明)
\[
\begin{align}
z_1z_2 &= (r_1\cos(\theta_1)+ir_1\sin(\theta_1))(r_2\cos(\theta_2)+ir_2\sin(\theta_2)) \\
\\
&\underbrace{=}_{積の公式(3)}r_1\cos(\theta_1)r_2\cos(\theta_2)-r_1\sin(\theta_1)r_2\sin(\theta_2)\\
\\& +i(r_1\cos(\theta_1)r_2\sin(\theta_2)+r_2\cos(\theta_2)r_1\sin(\theta_1)) \\
\\
&=r_1r_2\{\cos(\theta_1)\cos(\theta_2)-\sin(\theta_1)\sin(\theta_2)\\
\\
& +i(\cos(\theta_1)\sin(\theta_2)+\cos(\theta_2)\sin(\theta_1))\} \\
\\
&\underbrace{=}_{加法定理}r_1r_2\{\cos(\theta_1+\theta_2)+i(\sin(\theta_1+\theta_2))\}
\end{align}
\]
●\(z_1=r_1(\cos(\theta_1)+i\sin(\theta_1)), z_2=r_2(\cos(\theta_2)+i\sin(\theta_2))\) のとき
\[
\frac{z_2}{z_1} = \frac{r_2}{r_1}\{\cos(\theta_2-\theta_1)+i\sin(\theta_2-\theta_1)\}\;(z_1\neq 0) \tag{13}
\]
を証明せよ。
(証明)
\[
\begin{align}
\frac{z_2}{z_1} &= \frac{r_2(\cos(\theta_2)+i\sin(\theta_2))}{r1(\cos(\theta_1)+i\sin(\theta_1))} \\
\\
&\underbrace{=}_{商の公式(4)}\frac{\cancel{r_1}\cos(\theta_1)r_2\cos(\theta_2)+\cancel{r_1}\sin(\theta_1)r_2\sin(\theta_2)}{r_1^{\cancel{2}}\cos^2(\theta_1)+r_1^{\cancel{2}}\sin^2(\theta_1)}\\
\\
& +i\frac{\cancel{r_1}\cos(\theta_1)r_2\sin(\theta_2)-r_2\cos(\theta_2)\cancel{r_1}\sin(\theta_1)}{r_1^{\cancel{2}}\cos^2(\theta_1)+r_1^{\cancel{2}}\sin^2(\theta_1)}\\
\\
&\underbrace{=}_{加法定理}\frac{r_2}{r_1}(\cos(\theta_2-\theta_1)+i\sin(\theta_2-\theta_1))
\end{align}
\]