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$-1$．