How to find the roots of $(w−1)^4 +(w−1)^3 +(w−1)^2 +w=0$

Question

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 $k=0,1,2,3,4$

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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?

Answer 1

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.

Answer 2

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