2.
設複數\(\displaystyle z=cos\frac{2\pi}{17}+i sin\frac{2\pi}{17}\)，\(\overline{z}\)為\(z\)的共軛複數。若定義函數\(\displaystyle f(x)=\frac{\displaystyle \sum_{k=1}^{100}x^k+1+z}{x^{101}}=\frac{x^{101}+x^{99}+\ldots+x+1+z}{x^{101}}\)，則\(f(1+\overline{z})\)為　　　。
[解答]
\(\displaystyle f(x)=\frac{x^{100}+x^{99}+\cdots +x+1+z}{x^{101}-1+1} \)
\(\displaystyle      =\frac{x^{100}+x^{99}+\cdots +x+1+z}{(x-1)(x^{100}+x^{99}+\cdots +x+1)+1} \)
令 \( g(x)=x^{100}+x^{99}+\cdots +x+1 \)
所以，\(\displaystyle f(1+\bar{z})=\frac{g(1+\bar{z})+z}{\bar{z}g(1+\bar{z})+1}=z=cos\frac{2\pi }{17}+i sin\frac{2\pi }{17} \)

分子分母同乘以\(z\) ，\(z \times \bar{z} =1 \)。