Applying Rouché's Theorem

Question

Determine how many zeros of the following polynomial lie inside the circle $|z|=2$ \begin{equation} z^{5}+2z^{4}+z^{3}+20z^{2}+3z-1=0\end{equation}

My Reasoning

If we put $f(z)=z^{5}+2z^{4}$ and $h(z)=z^{3}+20z^{2}+3z-1$, then we see that for $|z|=2$, $|f(z)|\leq |z^{5}|+2|z^{4}|=32+32=64$ and $|h(z)|\leq|z^{3}|+20|z^{2}|+3|z|-1=8+40+12-1=59$. Hence, $|h(z)|<|f(z)|$ for $|z|=2$ and, using Rouché's theorem, we conclude that $f(z)+h(z) z^{5}+2z^{4}+z^{3}+20z^{2}+3z-1$ has the same number of zeroes inside the circle $|z|=2$ as $f(z)=z^{5}+2z^{4}$. That is $5$ zeroes.

First of all, is my suggested solution correct? If yes, then do you know other ways one could tackle this type of problem? Cheers!

Answer 1

Consider $f(z)=z^{5}+2z^{4}+z^{3}+20z^{2}+3z-1$ , $g(z)=20z^2$

For $|z|=2$ , $|f(z)-g(z)|\le79<80=|g(z)|$ apply Rouche's theorem.

Answer 2

No. Compare $f(-2)$ and $h(-2)$.

But if we have $|h(z)|<|f(z)|$ how you speak about number of $f(z) + h(z)$ roots?

Answer 3

Your $f$ has a zero on $|z|=2$, so Rouche does not apply. Rouche states:

Assume $f$,$h$ are meromorphic in $G$, and $B(a;r)\subset G$. If $f$ has no zeros nor poles on $|z-a|=r$ and $|h|,|f|$ on $|z-a|=r$, then $$Z_f-P_f = Z_{f+h}-P_{f+h}$$ where $Z_f$ and $P_f$ are the number of zeros and poles of $f$, respectively, that lie inside $B(a;r)$.

Also, to verify that $|f|>|h|$ on $|z-a|=r$ you should find a lower bound for $|f|$.

Try with $f(z)=z^5+20z^2$ and $h(z)=2z^4+z^3+3z-1$. You should get that the number or roots inside $B(0;2)$ is 2.