Show that all the roots of $p(z)= z^5 – z^3 + 1$ are in the ring $\frac{1}{2}\lt \vert z \vert \lt \frac{3}{2}$.

Question

I am going to try check that $\vert z^5 – z^3\vert \lt 1$ for $\vert z \vert \le \frac{1}{2}.$

$1=|-1|=|z^5+z^3|=|z|^3\cdot |z^2+1|\le |z^2+1|\le |z|^2+|1|=|z|^2+1 \le 1.$

Now I'm stuck. What do I do next?

Thank's for any help.

Answer 1

Using the triangle inequality, we get: $|z^5-z^3|=|z^3||z^2-1|\le\dfrac 18|z^2-1|\le\dfrac 18(|z^2|+1)\le\dfrac 18\left(\dfrac 14+1\right)=\dfrac 18\cdot \dfrac 54=\dfrac 5{32}\lt1$ for $|z|\le\dfrac 12$

Now let $g(z)=z^5$. Since $|z^3-1|\le|z^3|+1\le\dfrac {27}8+1=\dfrac {35}8\lt|z^5|=\left(\dfrac 32\right)^5$, on $|z|=\dfrac 32$, by Rouche we get that there are $5$ zeros in the given annulus.