I'd like to show $P(z)=1-2z^2-2z^3-2z^4-2z^5$ has a unique root inside the disk $|z|<0.6$.
I tried using Rouche's theorem, which worked for polynomials of this form $1-2z^2-2z^3-2z^4…-2z^n$ but of higher degrees, but the same method did not work on $n=5$ or smaller $n$.
For higher degrees, I used it as follows, multiplying by $z-1$ we obtain the polynomial $-2z^{n+1}+2z^2+z-1$.
By defining $f(z)=2z^2+z-1=2(z+1)(z-\frac{1}{2})$ and $g(z)=-2z^{n+1}$ one can show (using the regular and reverse triangle inequality) that on $|z|=0.6$ we have $|f(z)|>|g(z)|$ for $n\geq6$. However for $n=3,4,5$ this method failed.
I'm wondering if there's another way besides Rouche's theorem, or maybe a different use of Rouche here, or even some idea on why Rouche doesn't work on those values.
Just note that $0.6$ is not especially important, it's what I thought to use myself to find a proof of a certain claim. Similar radii (not far from $0.6$) that guarantee a unique root inside it would also be helpful.
Best Answer
I will expand on my comment. Following the OP's attempt, I will minimize $$F(x,y)=\big|f(x+yi)\big|=2\sqrt{\big((x+1)^2+y^2\big)\big((x-1/2)^2+y^2\big)}$$ subject to $x^2+y^2=r^2$ ($r$ is a non-negative constant). Let $$\mathcal{L}(x,y,\lambda)=\frac14\big(F(x,y)\big)^2+\lambda(x^2+y^2-r^2).$$ We set $$0=\frac{\partial \mathcal{L}}{\partial x}=2(x+1)\big((x-1/2)^2+y^2\big)+2(x-1/2)\big((x+1)^2+y^2)+2\lambda x,\tag{1}$$ $$0=\frac{\partial \mathcal{L}}{\partial y}=2y\big((x-1/2)^2+y^2\big)+2y\big((x+1)^2+y^2)+2\lambda y.$$ For the second equation, we have either that $y=0$ or $$(x-1/2)^2+(x+1)^2+2y^2+\lambda=0.\tag{2}$$ We also have the constraint condition $$y^2=r^2-x^2.\tag{3}$$ Thus $y=0$ yields solutions $$(x,y)=(\pm r,0).$$ We have $$a(r)=F(r,0)=|2r^2+r-1|$$ and $$b(r)=F(-r,0)=|2r^2-r-1|$$
From now on suppose that $y\ne 0$. Therefore $(2)$ holds. Plug $(3)$ into $(1)$ and $(2)$ to get $$(x+1)\big((x-1/2)^2-x^2+r^2\big)+(x-1/2)\big((x+1)^2-x^2+r^2\big)+\lambda x=0\tag{4}$$ and $$(x-1/2)^2+(x+1)^2-2x^2+2r^2+\lambda=0.$$ The previous equation gives $$x=-2r^2-\lambda-5/4.\tag{5}$$ Plug $(5)$ into $(4)$ to get $$-\left(2r^2+\lambda+\frac14\right)\left(3r^2+\lambda+\frac32\right)+\left(2r^2+\lambda+\frac74\right)\left(3r^2+2\lambda+\frac32\right)-\lambda\left(2r^2+\lambda+\frac54\right)=0.$$ That is, $$\lambda=-\frac{9(2r^2+1)}{8}.$$ This means $$x=-2r^2+\frac{9(2r^2+1)}{8}-\frac{5}{4}=\frac{2r^2-1}{8}.$$ Therefore $$y=\pm\frac{\sqrt{-4r^4+68r^2-1}}{8},$$ which is real only if $$0.12132\approx \frac{3\sqrt{2}-4}{2}\le r \le \frac{3\sqrt{2}+4}{2}\approx 4.12132.$$ Observe that \begin{align}c(r)&=F\left(\frac{2r^2-1}{8},\pm\frac{\sqrt{-4r^2+68r-1}}{8}\right)\\&=2\sqrt{\left(r^2+2\cdot\frac{2r^2-1}{8}+1\right)\left(r^2-\frac{2r^2-1}{8}+\frac14\right)}\\&=\frac{3(2r^2+1)}{2\sqrt2}.\end{align} We have $$\big(c(r)\big)^2-\big(a(r)\big)^2=\frac{(2r^2-8r-1)^2}{8}\geq 0$$ and $$\big(c(r)\big)^2-\big(b(r)\big)^2=\frac{(2r^2+8r-1)^2}{8}\geq 0.$$ Therefore, $a(r)\leq c(r)$ and $b(r)\leq c(r)$ always.
Therefore, the minimum of $F(x,y)$ with $x^2+y^2=r^2$ is $$m(r)=\min\{a(r),b(r)\}=\min\big\{|2r^2+r-1|,|2r^2-r-1|\big\}.$$ Since $m(0.6)=0.32$, we see that $$\big|g(z)\big|=\big|-2z^{n+1}\big|=2\cdot 0.6^{n+1}\leq 2\cdot 0.6^4=0.2592<0.32\leq \big|f(z)\big|$$ for $n\geq 3$ and $|z|=0.6$. Because $f(z)=2z^2+z-1$ has exactly one root $z=1/2$ inside the disk $|z|<0.6$, by Rouche's theorem, $$1-2z^2-2z^3-\ldots-2z^n=\frac{f(z)+g(z)}{z-1}$$ has exactly one root inside $|z|<0.6$. (If you replace $0.6$ by $0.7$, the statement is still true.)
Question: This left me wonder whether this is true. Let $f(z)$ be a polynomial function with only real roots. Is it true that the minimum value of $\big|f(z)\big|$ on any circle $|z|=r$ is attained at $z=r$ or $z=-r$ (clearly $z=\pm r$ are critical points)? Can someone prove or disprove this?
Edit: Fixed miscalculations.