Giải bài 28 trang 205 SGK giải tích nâng cao 12

Gợi ý: Số phức $z=a+bi$ có dạng lượng giác $z=r\left( \cos \varphi +i\sin \varphi \right)$, trong đó $\left\{ \begin{align} & r=\sqrt{{{a}^{2}}+{{b}^{2}}} \\ & \cos \varphi =\dfrac{a}{r},\sin \varphi =\dfrac{b}{r} \\ \end{align} \right.$

a) Ta có

$1-i\sqrt{3}=2\left( \dfrac{1}{2}-i\dfrac{\sqrt{3}}{2} \right)=2\left[ \cos \left( -\dfrac{\pi }{3} \right)+i\sin \left( -\dfrac{\pi }{3} \right) \right] \\ 1+i=\sqrt{2}\left( \dfrac{1}{\sqrt{2}}+i\dfrac{1}{\sqrt{2}} \right)=\sqrt{2}\left( \cos \dfrac{\pi }{4}+i\sin \dfrac{\pi }{4} \right) \\ $

Suy ra

$\begin{aligned} \left( 1-i\sqrt{3} \right)\left( 1+i \right)&=2\sqrt{2}\left[ \cos \left( -\dfrac{\pi }{3}+\dfrac{\pi }{4} \right)+i\sin \left( -\dfrac{\pi }{3}+\dfrac{\pi }{4} \right) \right] \\ & =2\sqrt{2}\left[ \cos \left( -\dfrac{\pi }{12} \right)+i\sin \left( -\dfrac{\pi }{12} \right) \right] \end{aligned}$

$\begin{aligned} \dfrac{1-i\sqrt{3}}{1+i}&=\sqrt{2}\left[ \cos \left( -\dfrac{\pi }{3}-\dfrac{\pi }{4} \right)+i\sin \left( -\dfrac{\pi }{3}-\dfrac{\pi }{4} \right) \right] \\ & =\sqrt{2}\left[ \cos \left( -\dfrac{7\pi }{12} \right)+i\sin \left( -\dfrac{7\pi }{12} \right) \right] \end{aligned}$

$\begin{aligned} b)\, 2i\left( \sqrt{3}-i \right)&=2+2i\sqrt{3}=4\left( \dfrac{1}{2}+i\dfrac{\sqrt{3}}{2} \right) \\ & =4\left[ \cos \left( \dfrac{\pi }{3} \right)+i\sin \left( \dfrac{\pi }{3} \right) \right] \\ \end{aligned}$

$\begin{align} c)\,\dfrac{1}{2+2i}&=\dfrac{1}{2}{{\left( 1+i \right)}^{-1}}\\ &=\dfrac{1}{2\sqrt{2}}\left[ \cos \left( -\dfrac{\pi }{4} \right)+i\sin \left( -\dfrac{\pi }{4} \right) \right] \end{align}$

d) $z=\sin \varphi +i\cos \varphi$

$\Rightarrow z=\cos \left( \dfrac{\pi }{2}-\varphi \right)+i\sin \left( \dfrac{\pi }{2}-\varphi \right)$

Ghi nhớ:

Nếu $z=r\left( \cos \varphi +i\sin \varphi \right) \\ z'=r'\left( \cos \varphi '+i\sin \varphi ' \right)\,\left( r\ge 0,r'\ge 0 \right)$

thì $zz'=rr'\left[ \cos \left( \varphi +\varphi ' \right)+i\sin \left( \varphi +\varphi ' \right) \right] \\ \dfrac{z}{z'}=\dfrac{r}{r'}\left[ \cos \left( \varphi -\varphi ' \right)+i\sin \left( \varphi -\varphi ' \right) \right]$

Công thức Moa-vrơ

${{\left[ r\left( \cos \varphi +i\sin \varphi \right) \right]}^{n}}={{r}^{n}}\left( \cos n\varphi +i\sin n\varphi \right)$