Show that all solutions of $z^5 - z + 16 = 0$ satisfy $1 \lt |z| \lt 2$

Question

I have to show that all solutions of $z^5 - z + 16 = 0$ satisfy $1 \lt |z| \lt 2$.

My attempt: By using Euler's formula I can rewrite the equation into $r^5e^{5i\phi} - re^{i\phi} = -16$ and then rewrite this into $$r^5 \cos(5\phi) - r \cos(\phi) + i(r^5 \sin(5\phi) -r \sin(\phi)) = -16 + i0 $$

so I get

1. $r^5 \cos(5\phi) - r \cos(\phi) = -16$
2. $r^5 \sin(5\phi) -r \sin(\phi) = 0$ which becomes $r^4 \sin(5\phi) = \sin(\phi)$

Here I don't know how to proceed. Any tips?

Answer 1

We have $|z^5-z|=|-16|=16.$

If $|z|=2$ then $|z^5-z|\ge |z^5|-|z|=|z|^5-2=30>16.$

If $|z|>2$ then $|z-1|\ge |z|-|1|>1$ and $|z|^4>0,$ so $|z^5-z|=|z|^4\cdot |z-1|> |z|^4\cdot 1 > 2^4=16.$

If $|z|\le 1$ then $|z^5-z|\le |z^5|+|z|=|z|^5+|z|\le 1^5+1=2<16.$

Answer 2

You get $|z| = \dfrac{16}{|z^4-1|}.$ If $|z|\leq1,$ then you have $16\leq|z^4-1|<|z|^4 +1 < 2,$ which is absurd. If $|z|\geq 2,$ then $8\geq |z^4-1|\geq||z|^4-1|\geq 15,$ which is also absurd. Looks like, this bound is not that tight.