Cho các số phức $z_{1}, z_{2}$ thỏa mãn: $\left | z_{1}-z_{2} \right |=\left | z_{1} \right |=\left | z_{2} \right |$>0. Hãy tính: A= $(\frac{z_{1}}{z_{2}})^{4}+(\frac{z_{2}}{z_{1}})^{4}$
Đặt $z_1=r(\cos\varphi _1+i\sin\varphi _1)$ ; $z_2=r(\cos\varphi _2+i\sin\varphi _2)$
Vì $\left | z_1-z_2 \right |=\left | z_1 \right |=\left | z_2 \right |\Rightarrow \left | \varphi _1-\varphi _2 \right |=\frac{\pi }{3}$
$\Rightarrow \left\{\begin{matrix}\frac{z_1}{z_2}=\frac{r}{r}(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3})=\frac{1}{2}+\frac{\sqrt{3}}{2}\ i\\\frac{z_2}{z_1}=\frac{r}{r}\left [ \cos\frac{-\pi}{3}+i\sin\frac{-\pi}{3} \right ] =\frac{1}{2}-\frac{\sqrt{3}}{2}\ i\end{matrix}\right.$
HOẶC
$\left\{\begin{matrix}\frac{z_1}{z_2}=\frac{r}{r}(\cos\frac{-\pi}{3}+i\sin\frac{-\pi}{3})=\frac{1}{2}-\frac{\sqrt{3}}{2}\ i\\\frac{z_2}{z_1}=\frac{r}{r}\left [ \cos\frac{\pi}{3}+i\sin\frac{\pi}{3} \right ]=\frac{1}{2}+\frac{\sqrt{3}}{2}\ i \end{matrix}\right.$
$\Rightarrow \left ( \frac{z_1}{z_2} \right )^4+\left ( \frac{z_2}{z_1} \right )^4=\left ( \frac{1}{2}+\frac{\sqrt{3}}{2}\ i \right )^4+\left ( \frac{1}{2}-\frac{\sqrt{3}}{2}\ i \right )^4$
$=\left ( -\frac{1}{2}-\frac{\sqrt{3}}{2}\ i \right )+\left ( -\frac{1}{2}+\frac{\sqrt{3}}{2}\ i \right )=-1$
Bài viết đã được chỉnh sửa nội dung bởi chanhquocnghiem: 01-12-2016 - 11:48