由 $z =\cos \dfrac{4\pi}{7} +\mathrm i \sin \dfrac{4\pi}{7}$ 知 $z$ 满足方程 $z^7 -1 =0$．又由 $z^7 -1 =(z - 1)(z^6 +z^5 +z^4 +z^3 +z^2 +z +1)$ 和 $z \ne 1$ 知 $z^6+z^5 +z^4 +z^3 +z^2 +z +1 = 0$．而 $\dfrac{z}{1+z^2} + \dfrac{z^2}{1+z^4} +\dfrac{z^3}{1+z^6}$ 通分后，分母为\[\begin{split} (1+z^2)(1+z^4)(1+z^6) &=1+z^2+z^4+z^6+z^8+z^{10}+z^{12} \\ &=1+z^2+z^4+z^6+z^6+z^{1}+z^{3} +z^5 \\& = z^6 ,\end{split}\]分子为\[\begin{split} z(1+z^4) (1+z^6) + z^2(1+z^2) (1+z^6) + z^3 (1+z^2) (1+z^4) & =(1+z^4)(1+z) + (1+z^2)(z^2 +z) + (1+z^2)(z^3+1) \\ & = 1+z +z^4 +z^5 +z +z^2 +z^3 +z^4 + 1 + z^2 +z^3 +z^5 \\&=2(1+z+z^2+z^3+z^4+z^5) \\&=-2z^6 .\end{split}\]于是$$\dfrac{z}{1+z^2} +\dfrac{z^2}{1+z^4} +\dfrac{z^3}{1+z^6} = \dfrac{-2z^6}{z^6} = -2.$$故 $\left \lvert \dfrac{z}{1+z^2}+ \dfrac{z^2}{1+z^4} +\dfrac{z^3}{1+z^6} \right \rvert$ 的值为 $2$．