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.


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

  • 1
    $\begingroup$ 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

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 $|z|>1$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.