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

What I've done so far:

$|z^4+z+3| \le |z|^4+|z|+3$.

If $0<|z|\le1, $ then $3< |z|^4+|z|+3 \le 5$.

How do I then complete the proof?

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    $\begingroup$ Hint: take norm and use triangle inequality. $\endgroup$ – edm Aug 6 '17 at 7:52
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    $\begingroup$ @edm $|z^4+z+3| \le |z|^4+|z|+3$. If $0<|z|\le1, $ then $3< |z|^4+|z|+3 \le 5$. How do I then complete the proof? $\endgroup$ – user538762 Aug 6 '17 at 7:58
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    $\begingroup$ More hint: you would want to use reverse triangle inequality. $\endgroup$ – edm Aug 6 '17 at 8:03
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    $\begingroup$ The first equation is equivalent to $z^4+z=-3$. $\endgroup$ – Angina Seng Aug 6 '17 at 8:09
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    $\begingroup$ You want to show that $|z^4+z+3|>0$. That means you should try to find something between $|z^4+z+3|$ and $0$. Finding something which is larger than $|z^4+z+3|$ doesn't help. $\endgroup$ – Arthur Aug 6 '17 at 8:12
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If $|z|\leq 1$ then $|z^4+z|= |z||z^3+1|\leq|z^3+1|\leq |z|^3+1\leq 1+1=2$.

So, if $|z|\leq 1$ then $|z^4+z|\leq 2$ and hence $z^4+z+3\neq 0$.

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  • $\begingroup$ Or just $|z^4+z|\le|z|^4+1\le1+1=2.$ $\endgroup$ – bof Aug 6 '17 at 8:54
  • $\begingroup$ Directly $|z^4+z|\leq |z|^4+|z|\leq 2$. Perfect. $\endgroup$ – Rafael Gonzalez Lopez Aug 6 '17 at 9:10
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Some comments point to the simplest answer. If $z^4+z+3=0$ then |z^4+z|=3. But if $|z|\le 1$ then $|z^4+z|\le |z^4|+|z|=|z|^4+|z|\le 2$, a contradiction.

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