Prove equality of two numbers written in complex polar form. Show that these two numbers are equal: 
$$
z_1=\frac{e^{\tfrac{2\pi i}{9}}-e^{\tfrac{5\pi i}{9}}}{1-e^{\tfrac{7\pi i}{9}}}
$$
and
$$z_2=\frac{e^{\tfrac{\pi i}{9}}-e^{\tfrac{3\pi i}{9}}}{1-e^{\tfrac{4\pi i}{9}}}=\frac{1}{e^{\tfrac{\pi i}{9}}+e^{\tfrac{-\pi i}{9}}}.
$$
In case it helps, I do know that they are both real.
Thanks in advance for any suggestions!
 A: Both complex numbers are of the form:$$\frac{e^{ia}-e^{ib}}{1-e^{i(a+b)}}=\frac{e^{ia}-e^{ib}}{1-e^{i(a+b)}}\times\frac{1+e^{-i(a+b)}}{1+e^{-i(a+b)}}=\frac{(e^{ia}-e^{-ia})-(e^{ib}-e^{-ib})}{-e^{i(a+b)}+e^{-i(a+b)}}\tag{1}$$We know that:$$\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$$$$\therefore2i\sin(x)=e^{ix}-e^{-ix}\tag{2}$$Using (2) in (1) we get:$$\frac{e^{ia}-e^{ib}}{1-e^{i(a+b)}}=\frac{2i\sin(a)-2i\sin(b)}{-2i\sin(a+b)}=\frac{\sin(b)-\sin(a)}{\sin(a+b)}=\frac{\sin(b)-\sin(a)}{2\sin(\frac{a+b}{2})\cos(\frac{a+b}{2})}\tag{3}$$Then, using Prosthaphaeresis Formulas in (3) we get:$$\require{cancel}\frac{e^{ia}-e^{ib}}{1-e^{i(a+b)}}=\frac{\cancel{2\cos(\frac{b+a}{2})}\sin(\frac{b-a}{2})}{\cancel{2}\sin(\frac{a+b}{2})\cancel{\cos(\frac{a+b}{2})}}=\frac{\sin(\frac{b-a}{2})}{\sin(\frac{a+b}{2})}\tag{4}$$If we now use (4) and the fact that $\sin(x)=\cos(\frac{\pi}{2}-x)$ with the values you have for $z_1$ and $z_2$ we get:$$z_1=\frac{\sin(\frac{3\pi}{18})}{\sin(\frac{7\pi}{18})}=\frac{\sin(\frac{3\pi}{18})}{\cos(\frac{2\pi}{18})}=\frac{1}{2\cos(\frac{2\pi}{18})}$$$$z_2=\frac{\sin(\frac{2\pi}{18})}{\sin(\frac{4\pi}{18})}=\frac{\cancel{\sin(\frac{2\pi}{18})}}{2\cancel{\sin(\frac{2\pi}{18})}\cos(\frac{2\pi}{18})}=\frac{1}{2\cos(\frac{2\pi}{18})}$$Hence:$$z_1=z_2$$
A: Let $w = \exp(i\pi/9)$. Then
$$
z_1 = \frac{w^2-w^5}{1-w^7} = \frac{w^2(1-w^3)}{1-w^7}
$$
and
$$
z_2 = \frac{w-w^3}{1-w^4} = \frac{w(1-w^2)}{(1-w^2)(1+w^2)} = \frac{w}{1+w^2}.
$$
Hence
$$
z_1 - z_2 = \frac{w^2(1-w^3)(1+w^2)-w(1-w^7)}{(1-w^7)(1+w^2)} =
\frac{w(1-w)(w^6-w^3+1)}{(1-w^7)(1+w^2)}.
$$ 
It remains to show that $w^6 - w^3 + 1 = 0$, but
$$
w^6 - w^3 + 1 = w^{-3}(w^9 - w^6 + w^3) = w^{-3}(1-w^6+w^3)
$$
which establishes that. There are probably shorter ways to see this.
