In preparation for qualifying exams I am working through old exams and came across the following question:
Determine the number of roots, counted with multiplicity, of the equation $$2z^5-15z^2+z+2$$ inside the annulus $1\leq |z|\leq 2$.
It seemed like a relatively straight forward application of Rouche's Theorem and was able to show there are two roots inside the unit disk, but when I was considering the boundary of $D(0,2)$, I couldn't seem to get a strict inequality in order to apply Rouche's Theorem. For example I chose $$f(z)=-15z^2+z+2$$ and $$g(z)=2z^5$$ but the best I could do was $|f(z)|\leq 64 =|g(z)|$ on $\partial D(0,2)$. Similar problems happened on different choices for $f$ and $g$.
P.S. I am trying to use: If $|f|<|g|$ on $\partial D(0,r)$ then $|Z_{D(0,r)}(g-f)|=|Z_{D(0,r)}g|$
Best Answer
Rouché's theorem is typically about the interior of a region (although the conditions on the boundary prevent roots there too). In your case, you're interested in $|z|\leq 2$. So, let's take your splitting and use $|z|=2+\varepsilon$ for some $\varepsilon>0$. In this case, $$ |f(z)|\leq 15(2+\varepsilon)^2+(2+\varepsilon)+2=64+61\varepsilon+\varepsilon^2 $$ and $$ |g(z)|=2(2+\varepsilon)^5=64+160\varepsilon+160\varepsilon^2+80\varepsilon^3+20\varepsilon^4+2\varepsilon^5. $$ Therefore, on the circle of radius $2+\varepsilon$, $|f(z)|<|g(z)|$, so there are the same number of roots of both $f$ and $g$ in the open disk $B(0,2+\varepsilon)$. But, this is true for every $\varepsilon$, so there can be no roots of $f$ outside the disk of radius $2$ (details hidden below).
For the disk of radius $1$, the $-15z^2$ should be a direct application of Rouche's theorem.