每日一题[732]复数的三角形式

已知两个非零复数$x,y$的立方和为$0$,则$\left(\dfrac{x}{x-y}\right)^{2000}+\left(\dfrac{y}{y-x}\right)^{2000}$的值为______.


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分析与解 $\dfrac{1}{2^{1999}}$或$-1$.

根据题意,有$x^3+y^3=0$,于是$$(x+y)\left(x^2-xy+y^2\right)=0.$$若$x+y=0$,则$$\left(\dfrac{x}{x-y}\right)^{2000}=\left(\dfrac{y}{y-x}\right)^{2000}=\dfrac{1}{2^{2000}},$$于是原式的值为$\dfrac{1}{2^{1999}}$.

若$x^2-xy+y^2=0$,设$z=\dfrac{x}{y}$,则$$z^2-z+1=0,$$进而$z=\cos\dfrac{\pi}3+{\rm i}\sin\dfrac{\pi}3$或$\overline{z}=\cos\dfrac{\pi}3+{\rm i}\sin\dfrac{\pi}3$,进而\[\begin{split} \left(\dfrac{x}{x-y}\right)^{2000}+\left(\dfrac{y}{y-x}\right)^{2000}&=\left(\dfrac{z}{z-1}\right)^{2000}+\left(\dfrac{1}{1-z}\right)^{2000}\\&=\left(\dfrac{z^2}{z^2-z}\right)^{2000}+\left(\dfrac{1}{-z^2}\right)^{2000}\\&=z^{4000}+{\overline z}^{4000}\\&=2\cos\dfrac{4000\pi}3\\&=-1.\end{split} \]综上所述,所求代数式的值为$\dfrac{1}{2^{1999}}$或$-1$.

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