$$(\frac{z+1}{z})^5=1$$ Show that its roots are $$-\frac{1}{2}(1+i\cot(\frac{kπ}{5})), k = 1,2,3,4$$ I need to use the five fifth roots of unit, with angles $0,\frac{π}{5}, \frac{2π}{5},\frac{3π}{5},\frac{4π}{5}$
I started by doing $$\frac{z+1}{z}=(\text{cis}(2πk))^{\frac{1}{5}}$$ $$z(1-\text{cis}(\frac{2πk}{5}))=-1$$ $$z=\frac{-1}{1-\text{cis}(\frac{2πk}{5})}$$ I am tempted to change the $1$ to cis form too, but that would eliminate the denominator and divide by $0$. How do I continue?
Hint: $$-{1 \over 1 - e^{2\pi i \over 5}} = {e^{-{\pi i \over 5}} \over e^{\pi i \over 5} -e^{-{\pi i \over 5}}}$$ Recognize the denominator in terms of another function?