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

Question

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

My working is as such: Let $|z| \leq 1$. Then consider $|z^4+z| \leq |z^4| + |z| \leq 1 + 1 = 2$.

So, $|z^4 + z| \leq 2$.

Thus $-2 \leq z^4 + z \leq 2$. So $1 \leq z^4+z+3 \leq 5$.

And so $z^4+z+3 \neq 0$.

Hence if $|z| \leq 1$ there is no root $z$, and so any root $z$ must satisfy $|z| < 1$.

Is this correct?

Answer 1

This is very close to right. The one bit that is not correct is your claim that $-2 \le z^4+z \le 2$. It is not correct since $z^4+z$ is complex, and you can't use $\le$ on complex numbers^†.

But you don't actually need that:

A solution to $z^4 + z +3 = 0$ has $|z^4 + z| = 3$ (add -3 to both sides and take absolute value). You've shown that $|z| \le 1 \to |z^4 + z| \le 2$. These contradict, so no solution with $|z| \le 1$ exists.

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^† You could argue that since $z$ satisfies $z^4+z=3$, $z^4 + z$ must be real, where $\le$ is valid, but while technically OK, the argument feels a little circular to me.

Answer 2

It's better to use Rouché's theorem, with $f(z)=z^4+z$ and $g(z)=3$ then on $z=1$ we have $$|f(z)|=|z^4+z^3|\leq2<3$$ then the number of zeros $z^4+z+3$ and $3$ are the same.