# Showing $1-2z^2-2z^3-2z^4-2z^5$ has a unique root inside the disk of radius 0.6

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.

• Have you tried using winding numbers? Apr 4 '20 at 5:33
• A short explanation on winding numbers : math.stackexchange.com/q/3139148 Apr 4 '20 at 6:14
• 3Blue1Brown's video on winding numbers is also a great introduction to the subject. Apr 4 '20 at 6:37
• Maybe you could try to actually minimize $\big|f(z)\big|$ on $|z|=0.6$, instead of just using a weak lower bound $2\cdot (1-0.6)\cdot(0.6-0.5)$. It turns out that the minimum value is $0.32$ (see here). This means: Roche's thm works for $n=3,4,5$ as well. An application of Lagrange multipliers doesn't seem to be too difficult. Apr 4 '20 at 7:13

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.

• Thank you for the elaborate answer, though I did not understand how $\lambda=-\frac{9(2r^2+1)}{8}$ from the equation that preceeded it, can you elaborate? Apr 5 '20 at 4:14
• Just expand the whole thing. The $\lambda^2$ terms cancel, leaving only the linear terms and the constant terms. Apr 5 '20 at 4:33
• How did you get $-\lambda^2$? The first term would give $-\lambda^2$, the second term would give $2\lambda^2$, and the third $-\lambda^2$. Are you forgetting that one of the $\lambda$s in the second term has $2$ in front of it? Apr 5 '20 at 4:36
• I see my mistake. Thanks. Yes, you are right, I forgot the 2 infront. Apr 5 '20 at 4:38
• Thank you, Lagrange multipliers worked well. Apr 6 '20 at 3:07