Roots of a complex equation are outside the unit disc using the triangle inequality

Question

Problem: Use the triangle inequality to show that the roots of the complex equation $$z^4+z+4=0$$ has roots all outside the unit disc $|z|\le1$

My Thought Process: Clearly I need to use the triangle inequality and this would be a proof by contradiction. Assuming that $|z|\le1$, then the triangle inequality gives $|z^4+z|\le|z^4|+|z|\le2$ but I'm not sure where my contradiction would be.

Answer 1

If $z^4+z+4=0$, then $z^4+z=-4$, so you'd need $\lvert z^4+z \rvert = 4 $. But you've shown that $\lvert z^4+z \rvert \leq 2$ for $\lvert z \rvert \leq 1$.

Answer 2

Note that $z^4+z+4=0$ implies that $-4=z+z^4$, and, by the triangle inequality, $$4=|-4|=|z+z^4|\leq |z|+|z|^4$$ Hence if $|z|=r$ then $4\leq r+ r^4$. Is it possible that $r\leq 1$?

By the same inequality, it turns out that $r$ has to be greater than $1.28$ because $1.28+(1.28)^4\approx 3.96<4$.