浙江省赛

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第5题 已知虚数{z}满足{z^3+1=0},则{\left(\frac{z}{z-1}\right)^{2018}+\left(\frac{1}{z-1}\right)^{2018}=\underline{~~~~~~~~~~}}.

 

解: 方程{z^{3}=-1}有两个虚根:{z_{1}=\cos \frac{\pi}{3}+i\sin \frac{\pi}{3}=\frac{1}{2}+\frac{\sqrt{3}}{2}i},{z_{2}=\cos -\frac{\pi}{3}+i\sin -\frac{\pi}{3}=\frac{1}{2}-\frac{\sqrt{3}}{2}i=\frac{1}{z_1}}.

  • {z=z_{1}=\frac{1}{2}+\frac{\sqrt{3}}{2}i}时,

    \displaystyle \frac{z_1}{z_1-1}=\frac{\frac{1}{2}+\frac{\sqrt{3}}{2}i}{-\frac{1}{2}+\frac{\sqrt{3}}{2}i}=\frac{\cos \frac{\pi}{3}+i\sin \frac{\pi}{3}}{\cos \frac{2\pi}{3}+i\sin \frac{2\pi}{3}}=\cos -\frac{\pi}{3}+i\sin -\frac{\pi}{3}=\frac{1}{2}-\frac{\sqrt{3}}{2}i,

    \displaystyle \frac{1}{z_1-1}=\frac{z_1}{z_1-1}-1=-\frac{1}{2}-\frac{\sqrt{3}}{2}i=\cos -\frac{2\pi}{3}+i\sin -\frac{2\pi}{3}.

    所以

    \displaystyle \begin{array}{rcl} \left(\frac{z_1}{z_1-1}\right)^{2018}+\left(\frac{1}{z_1-1}\right)^{2018}&=&\left(\cos -\frac{\pi}{3}+i\sin -\frac{\pi}{3}\right)^{2018}+\left(\cos -\frac{2\pi}{3}+i\sin -\frac{2\pi}{3}\right)^{2018} \\&=&\left(\cos -\frac{2018}{3}\pi+i\sin -\frac{2018}{3}\pi\right)+\left(\cos -\frac{4036}{3}\pi+i\sin -\frac{4036}{3}\pi\right) \\&=&\left(\cos -\frac{2}{3}\pi+i\sin -\frac{2}{3}\pi\right)+\left(\cos -\frac{4}{3}\pi+i\sin -\frac{4}{3}\pi\right) \\&=&-1 \end{array}

  • {z=z_2=\frac{1}{z_1}}时,

    \displaystyle \left(\frac{z_2}{z_2-1}\right)^{2018}+\left(\frac{1}{z_2-1}\right)^{2018}=\left(\frac{\frac{1}{z_1}}{\frac{1}{z_1}-1}\right)^{2018}+\left(\frac{1}{\frac{1}{z_1}-1}\right)^{2018}=\left(\frac{1}{z_1-1}\right)^{2018}+\left(\frac{z_1}{z_1-1}\right)^{2018}=-1.

所以答案是{-1}. \Box

第9题 设{x,y\in \mathbf{R}}满足{x-6 \sqrt{y}-4 \sqrt{x-y}+12=0},则{x}的取值 范围为{\underline{~~~~~~~~~~}}.

解: 由柯西不等式,

\displaystyle x+12=6 \sqrt{y}+4 \sqrt{x-y}\leq \sqrt{6^{2}+4^{2}}\sqrt{(\sqrt{y})^2+(\sqrt{x-y})^{2}}=\sqrt{52x},

其中{x\geq y\geq 0},等号当且仅当{\frac{x}{y}=\frac{13}{9}}时取得.因此{(x+12)^2\leq 52x},解得{14-2 \sqrt{13} \leq x\leq 14+2 \sqrt{13}}. \Box

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