Giải bài 33 trang 207 SGK giải tích nâng cao 12
Tính
\({{\left( \sqrt{3}-i \right)}^{6}};\) \({{\left( \dfrac{i}{1+i} \right)}^{2004}};\) \({{\left( \dfrac{5+3i\sqrt{3}}{1-2i\sqrt{3}} \right)}^{21}}.\)
Ta có
\(\begin{aligned} +)\, \sqrt{3}-i&=2\left( \dfrac{\sqrt{3}}{2}-\dfrac{1}{2}i \right) \\ & =2\left[ \cos \left( -\dfrac{\pi }{6} \right)+i\sin \left( -\dfrac{\pi }{6} \right) \right] \\ \end{aligned} \)
\(\Rightarrow {{\left( \sqrt{3}-i \right)}^{6}}={{2}^{6}}\left[ \cos \left( -\pi \right)+i\sin \left( -\pi \right) \right]=-{{2}^{6}}\).
\(\begin{aligned} +)\, \dfrac{i}{1+i}&=\dfrac{i\left( 1-i \right)}{2}=\dfrac{1+i}{2} \\ & =\dfrac{\sqrt{2}}{2}\left( \cos \dfrac{\pi }{4}+i\sin \dfrac{\pi }{4} \right) \\ \end{aligned} \)
\(\begin{aligned} \Rightarrow {{\left( \dfrac{i}{1+i} \right)}^{2004}}&={{\left( \dfrac{\sqrt{2}}{2} \right)}^{2004}}{{\left( \cos \dfrac{\pi }{4}+i\sin \dfrac{\pi }{4} \right)}^{2004}} \\ & =\dfrac{1}{{{2}^{1002}}}\left( \cos \dfrac{2004\pi }{4}+i\sin \dfrac{2004\pi }{4} \right) \\ & =\dfrac{1}{{{2}^{1002}}}\left( \cos \pi +i\sin \pi \right) \\ & =-\dfrac{1}{{{2}^{1002}}} \\ \end{aligned} \)
\(\begin{aligned} +)\, \dfrac{5+3i\sqrt{3}}{1-2i\sqrt{3}}&=\dfrac{\left( 5+3i\sqrt{3} \right)\left( 1-2i\sqrt{3} \right)}{1-2i\sqrt{3}} \\ & =\dfrac{-13+13i\sqrt{3}}{13} \\ & =-1+i\sqrt{3} \\ \end{aligned} \)
\(\begin{aligned} \text{Mà}\, -1+i\sqrt{3}&=2\left( -\dfrac{1}{2}+i\dfrac{\sqrt{3}}{2} \right) \\ & =2\left( \cos \dfrac{2\pi }{3}+i\sin \dfrac{2\pi }{3} \right) \\ \end{aligned} \)
\(\begin{aligned} \Rightarrow {{\left( \dfrac{5+3i\sqrt{3}}{1-2i\sqrt{3}} \right)}^{21}}&={{2}^{21}}{{\left( \cos \dfrac{2\pi }{3}+i\sin \dfrac{2\pi }{3} \right)}^{21}} \\ & ={{2}^{21}}\left( \cos \dfrac{42\pi }{3}+i\sin \dfrac{42\pi }{3} \right) \\ & ={{2}^{21}}\left( \cos 14\pi +i\sin 14\pi \right) \\ & ={{2}^{21}} \\ \end{aligned} \)
Ghi nhớ:
Số phức \(z=a+bi\) có dạng lượng giác \(z=r\left( \cos \varphi +i\sin \varphi \right)\), trong đó \(\left\{ \begin{align} & r=\sqrt{{{a}^{2}}+{{b}^{2}}} \\ & \cos \varphi =\dfrac{a}{r},\sin \varphi =\dfrac{b}{r} \\ \end{align} \right. \)
Công thức Moa-vrơ
\({{\left[ r\left( \cos \varphi +i\sin \varphi \right) \right]}^{n}}={{r}^{n}}\left( \cos n\varphi +i\sin n\varphi \right)\)