Giải bài 4, 5 trang 189 SGK giải tích nâng cao 12

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4. Thực hiện phép tính

$\begin{align} & \dfrac{1}{2-3i}=\dfrac{2+3i}{4+9}=\dfrac{2}{13}+\dfrac{3}{13}i \\ & \dfrac{1}{\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}i}=\dfrac{\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i}{\dfrac{1}{4}+\dfrac{3}{4}}=\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i \\ & \dfrac{3-2i}{i}=\dfrac{\left( 3-2i \right)\left( -i \right)}{1}=-2-3i \\ & \dfrac{3-4i}{4-i}=\dfrac{\left( 3-4i \right)\left( 4+i \right)}{16+1}=\dfrac{16}{17}-\dfrac{13}{17}i \\ \end{align}$

5. Với $z=-\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i$ ta có

$\dfrac{1}{z}=\dfrac{1}{-\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i}=\dfrac{-\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}i}{\dfrac{1}{4}+\dfrac{3}{4}}=-\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}i \\ \overline{z}=-\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}i \\ {{z}^{2}}={{\left( -\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i \right)}^{2}}=\dfrac{1}{4}-\dfrac{\sqrt{3}}{2}i-\dfrac{3}{4}=-\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}i$

$\begin{align} {{\left( \overline{z} \right)}^{3}} &={{\left( -\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}i \right)}^{3}} \\ & =-\dfrac{1}{8}\left( 1+3\sqrt{3}i-9-3\sqrt{3}i \right) \\ & =1 \\ \end{align}\\ 1+z+{{z}^{2}}=1+-\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i+-\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}i=0$