How to find the roots of $(w−1)^4 +(w−1)^3 +(w−1)^2 +w=0$ 
Write down, in any form, all the roots of the equation $z^5 − 1 = 0$
Hence find all the roots of the equation
$$(w−1)^4 +(w−1)^3 +(w−1)^2 +w=0$$
and deduce that none of them is real

My Try:
I know how to do the first part:
$$z^5=1=cos 2\pi k + i sin 2\pi k$$
$$z= cos \frac{2\pi k}{5}+ i sin \frac{2\pi k}{5}$$
where $k=0,1,2,3,4$

Please help me to do the second part. Thanks.
Attempt:
$$z^5-1=(z-1)(z^4+z^3+z^2+1)=0$$
Have to find:
$$(w−1)^4 +(w−1)^3 +(w−1)^2 +w=0$$
Let $w=z+1$
$$z^4+z^3+z^2+z+1=0$$
Multiply by $(z-1)$ each side:
$$(z-1)(z^4+z^3+z^2+z+1)=0 (z-1)=0$$
$$z^5-1=(z-1)(z^4+z^3+z^2+z+1)=0$$
$$z^5-1=0$$ (back to Inital result)
Then
$$z=cos \frac{2\pi k}{5}+ i sin \frac{2\pi k}{5}$$
Since $w=z+1$ So $z=w-1$
$$w-1=cos \frac{2\pi k}{5}+ i sin \frac{2\pi k}{5}$$
$$w=cos \frac{2\pi k}{5}+ i sin \frac{2\pi k}{5}+1$$
This seems weird, is this correct?

$$y:=w-1 \iff y^4+y^3+y^2+y+1=0 \stackrel{\cdot (y-1)}{\iff} (y-1)(y^4+y^3+y^2+y+1)=0 \iff y^5-1=0  \;\forall y\neq 1$$
