Solving $(z+1)^5 = z$ ($z\in\mathbb{C}$)

Question

I want to find all complex solutions to $$(z+1)^5 = z$$ My hope was to write it in the form $\alpha^k = 1$ for some $\alpha\in\mathbb{C}$ and $k\in\mathbb{N}$, then calculate the $k$th root of unity. In this case, since $z\neq 0$ that reduces this to $$\biggr(\cfrac{z+1}{z^{1/5}}\biggr)^5 = 1$$ I set $z = re^{i\theta}$. With some algebra the best I could do was that $$(r^{4/5}\exp(4i\theta/5) + r^{-1/5}\exp(-i\theta/5))^5 = 1$$ Now here I'm stuck.