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Show that any root $z$ of $z^4 + z + 3 = 0$ satisfies $|z|>1$.

I don't see any obvious way to show this; or any good geometrical interpretation if there is any.
I tried to consider Vieta's formulae, but wasn't sure what to make use of it.
I know that the roots will come in conjugate pairs by the complex conjugate root theorem. I also tried to do something like:
$z^4 + z + 3 = 0 \implies z^2 + z^{-1} + 3z^{-2} = 0 \implies 2\Re (z) + z^{-1} + 2z^{-2} = 0$ (basically any sort of algebraic manipulation) but to no avail.

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Hint : If $|z|\leq 1$, then we have $$|z^4+z|\leq |z|^4+|z|\leq 2<|-3|.$$

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    $\begingroup$ Hmm, not sure how to make use of it. I showed that $3 \leq |z^4 + z| + 3 \leq |z^4| + |z| + 3 \leq 5$ so the quantity $|z^4| + |z|+3$ is bounded between $3$ and $5$ so is never zero. But this is only true if we took the absolute value of $z$'s. $\endgroup$ – Twenty-six colours Jul 5 '17 at 14:12
  • $\begingroup$ I've added a small part to the hint. $\endgroup$ – Arnaud D. Jul 5 '17 at 14:16
  • $\begingroup$ Ahhh, I think I got it. Is this correct then: $|z^4 + z| \leq |z^4| + |z| \leq 2 < |-3| = |z^4 + z|.$ So this is saying that if $|z|\leq 1$, then we have $|z^4| + |z| < |z^4 + z|$ which is untrue (contradicts the Triangle Inequality)? If so, could I have assumed a larger bound of $|z|$ say $|z| \leq 1.15$ so with the same steps, $|z|^4 + |z| \leq 2.899.. < |-3|$ to conclude that $|z| > 1.15$? $\endgroup$ – Twenty-six colours Jul 5 '17 at 14:33
  • $\begingroup$ You get the idea. And yes, you can replace the condition $|z|\leq 1$ by $|z|\leq \alpha$ for any $\alpha\geq 0 $ such that $\alpha^4+\alpha < 3$. $\endgroup$ – Arnaud D. Jul 5 '17 at 14:38
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Let's assume that there exist such a complex number, $z_0$, with $|z_0|\leq1$, which is a root of the given equation. So: $$z_0^4+z_0+3=0\Leftrightarrow z_0^4=-(z_0+3)$$ Since $$|z_0|\leq1\Rightarrow|z_0|^4\leq1\Rightarrow|z_0^4|\leq1$$ On the other hand, by the traingular inequality, we have: $$|z_0^4|=|-(z_0+3)|=|z_0+3|\geq|3-|z_0||\overset{|z_0|\leq1}{=}3-|z_0|\geq2$$ which contradicts to our assumption that $|z_0|\leq1$ and, hence, $|z|>1$ for every root of the given equation.

Generally, one can show, in exactly the same way, that, for every $k>2$, for all the roots of the equation $$z^4+z+k=0$$ it is true that $|z|>1$.

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