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

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?

• This looks like an exercise on Rouche's theorem. – Lord Shark the Unknown Feb 3 '18 at 6:27
• @LordSharktheUnknown I haven't studied this theorem yet – JakeHaming Feb 3 '18 at 6:29
• Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. – robjohn Feb 3 '18 at 6:52

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

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 $2$.

• how does it give |z|<2? – JakeHaming Feb 3 '18 at 6:34
• I added something. See now. – Michael Rozenberg Feb 3 '18 at 6:38

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, $|z|\lt2$.

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 $5$.

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 $\vert z\vert \leq2$.