The value of $\sum _{r=1}^8[\sin \left(\frac{2\pi r}{9}\right)+i\cos \left(\frac{2r\pi }{9}\right)]$ is

$$\begin{gathered} \sum _{r=0}^8[\sin \left(\frac{2\pi r}{9}\right)+i\cos \left(\frac{2\pi r}{9}\right)]-i \\ i\cos \theta +\sin \theta =i{e}^{-i\theta } \\ \sum _{r=0}^{8}i.e^{i\left(\frac{2\pi r}{9}\right)}-i=[i+i{e}^{i\frac{2\pi }{9}}+i{e}^{i\frac{4\pi }{9}}\mathrm{.........}]-i \\ =\left[i(1+e^{i\frac{2\pi }{9}}+e^{i\frac{4\pi }{9}}\mathrm{............})\right]-i \\ =i\left[\frac{1-e^{i\frac{2\pi }{9}\times 9}}{1-e^{i\frac{2\pi }{9}}}\right]-i \\ =0-i=-i \end{gathered}$$