Simplifying Complex Numbers in Exponential Form 
$(a)$ Suppose $p$ and $q$ are points on the unit circle such that the line through $p$ and $q$ intersects the real axis. Show that if $z$ is the point where this line intersects the real axis, then $z = \dfrac{p+q}{pq+1}$.
$(b)$ Let $P_1 P_2 \dotsb P_{18}$ be a regular $18$-gon. Show that $P_1 P_{10}$, $P_2 P_{13}$, and $P_3 P_{15}$ are concurrent.

I already proved part $(a)$, and used it to prove part $(b)$ by making the diameter of the $18$-gon one unit and defining $P_1 P_{10}$ as the real axis and then plugging both $P_2 P_{13}$ and $P_3 P_{15}$ into $p$ and $q$ to get
$$
z_{1} = \dfrac{\mathrm{e}^{\mathrm{i}\frac{\pi}{9}} + \mathrm{e}^{12\mathrm{i}\frac{\pi}{9}}}{\mathrm{e}^{\mathrm{i}\frac{\pi}{9}} \cdot \mathrm{e}^{12\mathrm{i}\frac{\pi}{9}} + 1}
$$
and
$$
z_{2} = \dfrac{\mathrm{e}^{2\mathrm{i}\frac{\pi}{9}} + \mathrm{e}^{14\mathrm{i}\frac{\pi}{9}}}{\mathrm{e}^{2\mathrm{i}\frac{\pi}{9}} \cdot \mathrm{e}^{14\mathrm{i}\frac{\pi}{9}} + 1} \;.
$$
Then, I set $z_{1}$ and $z_2$ equal and simplified to get
$$
0 = \mathrm{e}^{8\mathrm{i}\frac{\pi}{9}} - \mathrm{e}^{6\mathrm{i}\frac{\pi}{9}} - \mathrm{e}^{5\mathrm{i}\frac{\pi}{9}} + \mathrm{e}^{3\mathrm{i}\frac{\pi}{9}} + \mathrm{e}^{2\mathrm{i}\frac{\pi}{9}} - 1 \;.
$$
However, I can't seem to get past this part to get something like $0=0$ that proves part $(b)$. Any tips?

Let $x=e^{i\pi /9}.$ We have $x^9=-1.$ $$\text { We have }\quad x^3-x^6=x^{-3}(x^6-x^9)=x^{-3}(x^6+1)=x^{-3}+x^3.$$ $$\text { So }\quad x^2-x^5=x^{-1}(x^3-x^6)=x^{-1}(x^{-3}+x^3).$$ $$\text { We have } \quad x^8-1=x^{-1}(x^9-x)=x^{-1}(-1-x)=(1+x^{-1})(-1).$$ $$\text { Therefore }\quad x^8-x^6-x^5+x^3+x^2-1=$$ $$=(x^8-1)+(x^3-x^6)+(x^2-x^5)=$$ $$= (1+x^{-1})(-1)+(x^{-3}+     x^3)+x^{-1}(x^{-3}+x^3)=$$     $$=(1+x^{-1})(-1)+(1+x^{-1}) (x^{-3}+x^3)=$$      $$=(1+x^{-1})(-1+x^{-3}+x^3)=$$     $$=(1+x^{-1})\cdot x^{-3}\cdot (1-x^3+x^6)=$$  $$=(1+x^{-1})\cdot x^{-3}\cdot (1+x^9)(1+x^3)^{-1}=0$$ because $1+x^9=0.$
