Prove all the roots of a given polynomial lie in the disk with radius $2$

Question

I need to show that the roots of the polynomial $z^5-z^4+15$ are in the disk $|z|<2$. But how do I do this? Am I supposed to compute all the roots?

Answer 1

By the triangle inequality we obtain: $$|z|^5=|z^4-15|\leq|z|^4+15,$$ which gives $|z|<2.$

Because by https://en.wikipedia.org/wiki/Descartes%27_rule_of_signs

the equation $$x^5-x^4-15=0$$ has one positive root and since $2^5-2^4-15>0$ and $1^5-1^4-15<0$,

we see that the positive root of the last equation is later than $2$.

Answer 2

Note that $|-z^4+15|\leq |z|^4+15\leq 31<|z|^5$ when $|z|=2$, and so it follows from Rouche's theorem that $z^5-z^4+15$ and $z^5$ have the same number of zeros in the disk $|z|<2$, namely $5$.

Answer 3

Set $\gamma(z)=-z^4+15$ and $f(z)=z^5$. For $\vert z\vert=2\,$ we have $\vert \gamma(z)\vert \leq 2^4+15=31<\vert 2^5\vert=32=\vert f(z)\vert$. By Rouché's Theorem the number of roots of $f$ in $\vert z\vert<2\,$ $(=5)$ concides with the ones of $z^5-z^4+15\,$ in $\vert z\vert<2\,$. Therefore, $\,z^5-z^4+15\,$ has $5$ roots in $\vert z\vert<2\,$; being $5$ the total number of roots of $z^5-z^4+15\,$ in $\mathbb{C}$, then the given polynomial has all its roots in the disk $\vert z\vert \leq2$.

Answer 4

If $|z|\ge2$, then the triangle inequality says that $$ \begin{align} |z^5-z^4+15| &\ge|z|^4|z-1|-15\\[6pt] &\ge|z|^4(|z|-1)-15\\[6pt] &\ge16\cdot(2-1)-15\\[6pt] &=1 \end{align} $$ Therefore, $|z|\lt2$.