Roots of polynomial equation $x^6+2x^5+4x^4+8x^3+16x^2+32x+64=0$ 
If $x_1,x_2,...,x_6$ be the roots of $x^6+2x^5+4x^4+8x^3+16x^2+32x+64=0$ then I have to show that $|x_j|=2\space\space\space\forall j\in\{1,2,3,4,5,6\}$

I get that the roots of the equation must be complex, of the form $a+ib$ where $|a+ib|=\sqrt{a^2+b^2}=2$ for all $|x_i|$. I don't get how to find the value of $a+ib$. How do I find the roots?

$$x^6+2x^5+4x^4+8x^3+16x^2+32x+64=0$$
Let $z=x/2$, so
$$z^6+z^5+z^4+z^3+z^2+z+1=0$$
from $z \neq 1$ 
Then $\frac{1-z^7}{1-z}=0 $
so $1-z^7=0$
so $z$ is the $n$ th root of $7$.
from $x=2z$ ,so $|x|=|2z|$ so $|x|=|2z|=2$
