# Zeroes of $2z^5-15z^2+z+2$

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 $$|Z_{D(0,r)}(g-f)|=|Z_{D(0,r)}g|$$

• What happens if you use a circle of radius $2+\varepsilon$ and $1-\varepsilon$ and take the limit as $\varepsilon$ goes to $0$? Apr 26, 2020 at 20:24

Rouche'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 fore every epsilon, so there can be no roots of $$f$$ outside the disk of radius $$2$$ (details hidden below).
If there were a root of $$f$$ outside the disk of radius $$2$$, then let $$r$$ be the radius of this root. Consider two radii $$2+\varepsilon_1. Applying Rouche's theorem to both of these values gives that $$f$$ has the same number of roots within disks of radius $$2+\varepsilon_1$$ and $$2+\varepsilon_2$$. This, however, is impossible since $$f$$ has a least one fewer root within the smaller disk whereas $$g$$ does not.
For the disk of radius $$1$$, the $$-15z^2$$ should be a direct application of Rouche's theorem.
Actually your idea does work: With $$f(z)=-15z^2+z+2$$ and $$g(z)=2z^5$$ you have for $$|z| = 2$$ $$|f(z)| = |-15z^2+z+2| < |-15 z^2| + |z|+ 2 = 64 = |g(z)|.$$ Here we have strict inequality in the triangle inequality because the numbers $$-15z^2$$, $$z$$ and $$2$$ cannot all have the same argument.
So Rouche's theorem can be applied to conclude that $$f+g$$ and $$g$$ have the same number of zeros in $$D(0, 2)$$.