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?

Your approach looks good: $z^5-1=0$ has five solutions that are evenly distributed around the unit circle, as defined by your trigonometric solution.  These are the black points in the following figure:

As you say, the left hand side factors into
$$z^5-1 = (z-1)(z^4+z^3+z^2+z+1).$$
If we divide off the first order term and substitute $z=w-1$ into the remaining we get
$$z^4+z^3+z^2+z+1 = (w-1)^4 + (w-1)^3 + (w-1)^2 + (w-1) + 1.$$
Thus, the roots of the polynomial in $w$ are exactly the (non-real) roots of the polynomial shifted to the right, as shown in red.  Hence, they cannot be real.
