On solving $z^n = (1 + z)^n$ with root of unity trick. I know that this question has already been sorta asked in this post, but as my solving strategy differs a bit, I didn't find the said post helpful. This is the c part of the problem 64 from chapter 1.3. of Complex Analysis with Applications by Asmar.
So I am trying to solve the real and imaginary part of $z$ where $z^n = (1 + z)^n$, which are claimed to be $\frac{-1}{2}$ and $\frac{1}{2}\cot(k\pi/n)$, respectively. One way to do this is to note that $z^n \neq 0$, divide by $(1 + z)^n$ and to conclude that $\frac{z}{z + 1}$ is a root of unity, so that $z = (z + 1)\omega_k$ for some root of unity $\omega_k$.
So far I have tried to solve the components of $z$ by both writing $z$ as $z = \frac{\omega_k}{1 - \omega_k}$ and just inspecting the real and imaginary components of $z = (1 + z)\omega_k$. Let $\mathrm{Arg(\omega_k)} = \theta_k = \frac{2\pi k}{n}$. The former way gives me the seemingly unhelpful solution of $z = \frac{1}{\sqrt{2(1 - \cos(\theta_k)}}(\cos(\theta_k) - 1 + i\sin(\theta_k))$, and the latter approach yields eventually $\mathrm{Re}(z) = \frac{\sin(\theta_k) - \cos(\theta_k) + \sin(\theta_k)\cos(\theta_k)}{\cos(\theta_k) - 1 - \sin(\theta_k)\cos(\theta_k)}$, and I don't know how to manipulate it any further.
Is there any way of salvaging either of there approaches so that in the end we get the desired $\mathrm{Re}(z) = \frac{-1}{2}$?

You seem to have made a slight mistake with your first method. The denominator should be $2(1-\cos {\theta_k})$, without the square root.
For your second method, show the calculation as to how you obtained it.
