# Zeros of Polynomial Using Rouche's Theorem

Currently I studying analytic combinatorics and am looking at the OGF of integer compositions with summands restricted to the set $$\{1,...,r\}$$ for $$r\geq2$$: $$C^{\{1,...,r\}}(z) = \frac{1-z}{1-2z+z^{r+1}}.$$

In particular, $$z^{r+1} -2z + 1$$ has a unique zero $$\rho_r \in (\frac{1}{2},1)$$ of minimal magnitude so that by a later theorem the coefficients of the series expansion of the OGF are of the order $$\rho_r^{-n}$$. However, I am struggling to understand why we know such a zero exists. From complex analysis, I know that Rouche's theorem can be used to discern the amount and locations of the zeros to establish this. I tried using cases of odd and even $$r$$ to find functions to apply the theorem but am getting nowhere. How might Rouche's theorem or some other approach help us reach this fact?

• Are you trying to prove that $z^{r+1} -2z + 1$ has a unique zero for $z$ a real in $(0.5, 1)$, or for $z$ a complex with $|z| \in (0.5, 1)$? Nov 19, 2022 at 21:48

## 1 Answer

We have$$z^{r+1}-2z+1=z(z^r-1)-(z-1)\\ =[z^r+z^{r-1}+\ldots +z-1](z-1)$$ The expression in the square brackets takes the value $$r-1$$ at $$z=1$$ and negative value at $$z={1\over 2}.$$ Therefore it vanishes in the interval $$({1\over 2},1)$$ by the intermediate value theorem. On the other hand for $$|z|=1-\delta$$ we have $$2|z|=2(1-\delta)$$ and $$|z^{r+1}+1|\le (1-\delta)^{r+1}+1\le (1-\delta)^3+1\\ =2-3\delta +3\delta^2-\delta^3<2(1-\delta)$$ for $$\delta>0$$ small enough. Then by the Rouche theorem the function $$z^{r+1}-2z+1$$ vanishes exactly once in $$|z|<1.$$

In conclusion the function vanishes only once for $$|z|<1$$ and the root is located in the interval $$({1\over 2},1).$$