$\left[\frac{1+\sin \frac{2 \pi}{9}+i \cos \frac{2 \pi}{9}}{1+\sin \frac{2 \pi}{9}-i \cos \frac{2 \pi}{9}}\right]^n=1$
$\Rightarrow \frac{\left[\frac{1+\cos \left(\frac{\pi}{2}-\frac{2 \pi}{9}\right)+i \sin \left(\frac{\pi}{2}-\frac{2 \pi}{9}\right)}{1+\cos \left(\frac{\pi}{2}-\frac{2 \pi}{9}\right)-i \sin \left(\frac{\pi}{2}-\frac{2 \pi}{9}\right)}\right]^n}{n}=1$
$\begin{aligned} & \Rightarrow\left[\frac{1+\cos \frac{5 \pi}{18}+i \sin \frac{5 \pi}{18}}{1+\cos \frac{5 \pi}{18}-i \sin \frac{5 \pi}{18}}\right]=1 \\ & \Rightarrow\left[\frac{2 \cos ^2 \frac{5 \pi}{36}+i 2 \sin \frac{5 \pi}{36} \cos \frac{5 \pi}{36}}{2 \cos ^2 \frac{5 \pi}{36}-i 2 \sin \frac{5 \pi}{36} \cos \frac{5 \pi}{36}}\right]^n=1 \\ & \Rightarrow\left[\frac{\cos \frac{5 \pi}{36}+i \sin \frac{5 \pi}{36}}{\cos \frac{5 \pi}{36}-i \sin \frac{5 \pi}{36}}\right)^n=1 \Rightarrow\left(\frac{e^{i 5 \pi / 36}}{e^{-i 5 \pi / 36}}\right)^n=1 \\ & \Rightarrow e^{i 5 n \pi / 18}=1 \Rightarrow \frac{5 n \pi}{18}=2 \mathrm{~K} \pi, \mathrm{K} \in \mathrm{I} \Rightarrow n=\frac{36 \mathrm{~K}}{5}\end{aligned}$