$z^3 + z + 1 = 0,\quad z \in \mathbb C$ \ $\mathbb R, \quad$ prove $|z| \in (1, \sqrt2)$
$z = a + ib$
(I have the solution for this problem, but I don't understand the following part of it):
Now, I got $$a^2+b^2 > 1$$ But I don't understand how this implies $$z < 1$$ and then how do we get $$|z|^3 = |z+1|$$
Can someone explain this to me? (I don't need you to solve the entire problem for me, as I understand the other parts of the solution. But if you are interested or have time you can do it)
If you can show that a solution $z=a+bi$ to the equation $z^3+z+1=0$ must have the property that $a^2+b^2>1$ then you are implying that $|z|>1$ by definition of the absolute value of a complex number, not that $|z|<1$. But that's half of what we needed to show anyway, that $|z|\in(1,\sqrt{2})$.
To see that $|z^3|=|z+1|$ just rearrange your equation and take absolute values: $$ z^3+z+1=0 \quad \Leftrightarrow \quad z^3=-z-1 \quad \Rightarrow \quad |z^3|=|-z-1|=|z+1| $$