Умножение и деление комплексных чисел в тригонометрической форме.

Пусть $z_{1} = r_{1}(\cos\varphi_{1} + i\sin\varphi_{1}), z_{2} = r_{2}(\cos\varphi_{2} + i\sin\varphi_{2})$, тогда: $$z_{1}z_{2} = r_{1}r_{2}(\cos(\varphi_{1} + \varphi_{2}) + i\sin(\varphi_{1} + \varphi_{2}))$$ $$\dfrac{z_{1}}{z_{2}} = \dfrac{r_{1}}{r_{2}}(\cos(\varphi_{1} - \varphi_{2}) + i\sin(\varphi_{1} - \varphi_{2}))$$

Вычисляем напрямую: $$\begin{align} z_{1}z_{2} &= r_{1}r_{2}(\cos\varphi_{1}+i\sin\varphi_{1})(\cos\varphi_{2}+i\sin\varphi_{2}) = \\ &=r_{1}r_{2}[(\cos\varphi_{1}\cos\varphi_{2}-\sin\varphi_{1}\sin\varphi_{2})+i(\cos\varphi_{1}\sin\varphi_{2}+\sin\varphi_{1}\cos\varphi_{2})]= \\ &=r_{1}r_{2}(\cos(\varphi_{1}+\varphi_{2})+i\sin(\varphi_{1}+\varphi_{2})) \end{align}$$ $$\begin{align} \dfrac{z_{1}}{z_{2}} &= \dfrac{r_{1}(\cos\varphi_{1}+i\sin\varphi_{1})}{r_{2}(\cos\varphi_{2}+i\sin\varphi_{2})}=\dfrac{r_{1}(\cos\varphi_{1}+i\sin\varphi_{1})(\cos\varphi_{2}-i\sin\varphi_{2})}{r_{2}(\cos\varphi_{2}+i\sin\varphi_{2})(\cos\varphi_{2}-i\sin\varphi_{2})}= \\ &= \dfrac{r_{1}((\cos\varphi_{1}\cos\varphi_{2}+\sin\varphi_{1}\sin\varphi_{2})+i(\sin\varphi_{1}\cos\varphi_{2}-\cos\varphi_{1}\sin\varphi_{2}))}{r_{2}(\cos^{2}\varphi_{2}+\sin^{2}\varphi_{2})}= \\ &= \dfrac{r_{1}}{r_{2}}=(\cos(\varphi_{1}-\varphi_{2})+i\sin(\varphi_{1}-\varphi_{2})) \end{align}$$