Show that if $z$ is a complex number such that $z^5$ + 5$z^2$ + 7 = 0, then |$z$| > 1.

Question

How do I show that if $z$ is a complex number such that $z^5$ + 5$z^2$ + 7 = 0, then |$z$| > 1.

I thought maybe try by contradiction i.e letting |$z$| $\leq$ 1.

But then I ended up getting |$z^5$ + 5$z^2$ + 7| $\leq$ 13, which doesn't help my case.

Any ideas?

Answer 1

If $|z|\leq 1$ then $|z^{5}+5z^{2}+7| \geq 7-|z^{5}+5z^{2}|\geq 7-1-5=1$ so $z^{5}+5z^{2}+7$ cannot be $0$. [I have used the inequality $|a+b| \geq |a|-|b|$ which is true for all $a,b \in \mathbb C]$.

Answer 2

I'd first rewrite your equality: $$ -7=z^5+5z^2. $$ Now take absolute values, and use the triangle inequality: $$ 7\le |z|^5+5|z|^2. $$ Now argue by contradiction: if $|z|\le 1$, then the right hand side can be bounded by $$ 7\le 1^5+5\cdot 1^2=6, $$ which is clearly impossible.