How to find the roots of $(\frac{z-1}{z})^5=1$ 
Write down the fifth roots of unity in the form $\cos \theta + i \sin \theta$ where $ 0 \leq \theta \leq 2\pi$
Hence, or otherwise, find the fifth roots of i in a similar form
By writing the equation $(z-1)^5=z^5$ in the form :
$${\left(\frac{z-1}{z}\right)}^{5}=1$$
show that its roots are:
$$\frac{1}{2}\left(1- i \cot \frac{1}{4} \pi k\right)$$ where $k=0,1,2,3,4$

I know how to do the first 2 parts :
First part:
$$z^5=1= \cos 2\pi + i \sin 2 \pi $$
$$z= \left( \cos \frac{2\pi k}{5} + i \sin \frac{2\pi k}{5}\right)$$
where $k=0,1,2,3,4$

Second Part:
$$z^5=i= \cos \pi + i sin \pi $$
$$z= \left(\cos \frac{\pi k}{5} + i \sin \frac {\pi k}{5}\right)$$
where $k=0,1,2,3,4$

I have no clue how to do the third part, Please help.
How to do the third part?
Please note: I am a beginner to the complex world
 A: 
Solve$$\displaystyle \left(\frac{z-1}{z}\right)^5=1\rightarrow \frac{z-1}{z}=e^{\Large \frac{i\cdot{2k}\pi}{5}}$$ for  $k=0,1,2,3,4$. 

Hence $$\displaystyle 1-\frac{1}{z}=e^{\Large \frac{i\cdot{2k}\pi}{5}}\rightarrow 1-e^{\Large \frac{i\cdot{2k}\pi}{5}}=\frac{1}{z}\rightarrow z=\frac{1}{1-e^{\Large \frac{i\cdot{2k}\pi}{5}}}=\frac{1}{\left(1-\cos\left(\frac{2k\pi}{5}\right)\right)-i\sin\left(\frac{2k\pi}{5}\right)}$$ Now, $$\frac{\left(1-\cos\left(\frac{2k\pi}{5}\right)\right)+i\sin\left(\frac{2k\pi}{5}\right)}{\left(1-\cos\left(\frac{2k\pi}{5}\right)\right)^2+\sin^2\left(\frac{2k\pi}{5}\right)}=\frac{\left(1-\cos\left(\frac{2k\pi}{5}\right)\right)+i\sin\left(\frac{2k\pi}{5}\right)}{2-2\cos\left(\frac{2k\pi}{5}\right)}=\\=\frac{2\sin^2\left(\frac{k\pi}{5}\right)+i\cdot 2\sin\left(\frac{k\pi}{5}\right)\cos\left(\frac{k\pi}{5}\right)}{4\sin^2\left(\frac{k\pi}{5}\right)}=\frac{\sin\left(\frac{k\pi}{5}\right)+i\cos\left(\frac{k\pi}{5}\right)}{2\sin\left(\frac{k\pi}{5}\right)}=\\=\frac{1}{2}\left(1+i\cdot \cot\left(\frac{k\pi}{5}\right)\right)$$
So the question is incorrect. 
