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Question Statement:-

If $z$ is a complex number such that $z^4+z+2=0$, show that $z$ cannot lie in the interior of the circle $|z|=1$

Attempt at a solution :-

We are given with the equation $z^4+z+2=0$, so we can write

$$z^4+z+2=0\implies z^4+z=-2 \implies |z^4+z|=2$$

Now, from triangle inequality we get

$$|z^4+z|\le |z^4|+|z|\implies |z|^4 + |z|\ge 2$$

My deal with the question:-

Now I did think that this inequality can hold only when $|z|\ge 1$, but I don't know why but I don't feel very comfortable with this type of proof.

So, if anyone can provide me a rigorous algebraic approach to the question. And as always more elegant solutions are welcome.

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  • Why did you write $|z|^4+|z|\ge2$ out of nowhere? I don't see what justification you have to write that. Instead, use proof by contradiction. First assume $z$ is a solution with $|z|<1$. Then at the end conclude $|z^4+z|\le |z|^4+|z|<1+1=2$, contradicting $|z^4+z|=2$. – anon Jul 30 '16 at 06:13
  • @arctic: If you see carefully, in my solution I clearly have stated why $|z^4+z|=2$ in the very first line. – user350331 Jul 30 '16 at 06:16
  • If you see carefully, in my comment I clearly have asked why you wrote down $|z|^4+|z|\ge2$, not $|z^4+z|=2.~~$ :-) – anon Jul 30 '16 at 06:17
  • Oh...then also still it is correct due to the property I stated in my solution, the "Triangle inequality" – user350331 Jul 30 '16 at 06:19
  • You mean $2=|z^4+z|\le |z|^4+|z|$? (If you're just starting out with proofs, you'll want the equations you write down to clearly indicate which properties are being used.) Yes, that's correct, but where could you go from there? [Also, this question doesn't seem to have anything to do with algebraic geometry.] – anon Jul 30 '16 at 06:21
  • @arctictern: I did state the property I used in my solution, and about the tag I just thought that it relates with the solutions lying outside the circle(there a geometrical figure), so I tagged it because I thought that only one tag doesn't look good :-P. If its not appropriate I will remove it immediately. – user350331 Jul 30 '16 at 06:28

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If $|z| \lt 1$ then $|z|^4 \le |z| \lt 1$ therefore $|z|^4 + |z| \lt 1 + 1=2$ which contradicts the inequality $|z|^4 + |z|\ge 2$

  • Okay, that was really easy, why didn't I think of that, if possible can you also provide any other kind of solution that you can come up with. – user350331 Jul 30 '16 at 06:22
  • @user350331 Are you familiar with Rouché's theorem? –  Jul 30 '16 at 06:29