Prove that $2z^4-3z^3+3z^2-z+1=0$ has exactly one complex root in each of the four quadrants.

I am trying to show that $$p(z)=2z^4-3z^3+3z^2-z+1=0$$ only has a single root in all four quadrants.

From two previously related posts, I have shown that $$p(z)$$ does not have a root on neither the imaginary or real axes. We also know that as the coefficients of $$p(z)$$ are real, then the roots of $$p(z)$$ occur in complex conjugate pairs. Hence we have roots $$w_1=a\pm ib \ \ \text{and} \ \ w_2=c\pm id \ \ \text{where} \ \ a,b,c,d\in\mathbb{R} \ \ \text{and} \ \ a,b,c,d\neq 0.$$ Ideally if we could show that $$\Re(w_1)=-\Re(w_2)$$, then we could conclude the result. But the sum of the roots is $$\frac{3}{2}$$ and not $$0$$.

How can we show only a single root exists in each quadrant?

edit

In consultation with my professor, he suggested using a corollary of Cauchy's argument principle:

"If $$f\in H(\Omega)$$, $$\Omega$$ is a domain, $$\gamma:[a,b]\rightarrow\Omega$$ is a simple closed contour, $$f(\gamma(t))\neq 0 \ \ \forall t\in [a,b]$$, then the number of zeros of $$f$$ in Int($$\gamma$$) (counting multiplicities) is equal to the number of times $$f\circ\gamma$$ winds around $$0$$".

We can consider $$\gamma(t)=Re^{it}$$ where $$t\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$$ and $$R>0$$. We would expect the result of this to be $$2$$, meaning that there is a root $$w$$ in the first quadrant and $$\overline{w}$$ in the forth quadrant.

• math.stackexchange.com/questions/2967323/… math.stackexchange.com/questions/2967260/… You could have merged these together in single question instead of asking them as separate question. – A B Oct 23 '18 at 10:11
• The sum of the real parts of the roots is $3/4$ and not $3/2$. – Fabian Oct 23 '18 at 10:13
• @StammeringMathematician I was unsure of the correct protocol. I will do this in the future, apologies – M B Oct 23 '18 at 10:14
• @StammeringMathematician, OP is trying to solve exercise by themselves with a bit of help. That question was an attempt that didn't work. It's much better to ask a new question than to edit the old one over and over. Or try to seek an answer to altered question in comments (in those cases most of us will at some point say to ask a new question). As long as the streak of new questions on the same thing is not excessive (whatever that is), I think that's fine. – Ennar Oct 23 '18 at 10:52
• @StammeringMathematician if you read closely, that question was answered. It was further advised in the comment section that if the OP had any firther questions, they should be addressed in a new post. – user557493 Oct 23 '18 at 11:02

There is a slightly more straightforward way to prove the claim by the argument principle.

Since the coefficients of $$p(z)$$ are all real and the polynomial has no roots on axis, we know that there are two roots in the upper half plane and two roots in the lower half plane. Hence it suffices to show that exactly two roots occur in the right half plane.

Then we consider the half circle contour $$T_R=I_R \cup C_R$$ where $$I_R=\{ it:t \in [-R,R] \}$$ and $$C_R=\{ Re^{i\theta}: \theta \in [-\frac{\pi}{2}, \frac{\pi}{2}] \}$$. We would like to study the total change of argument of $$p(z)$$ along the contour $$T_R$$.

On $$C_R$$, we observe that $$\int_{C_R}\frac{p'(z)}{p(z)}\mathrm{d}z =\int_{-\pi/2}^{\pi/2} \frac{p'(Re^{i\theta})}{p(Re^{i\theta})} iRe^{i\theta} \mathrm{d}\theta =\int_{-\pi/2}^{\pi/2} \frac{8R^3e^{3i\theta}-9R^2e^{2i\theta}+6Re^{i\theta}-1} {2R^4e^{4i\theta}-3R^3e^{3i\theta}+3R^2e^{2i\theta}-Re^{i\theta}+1} iRe^{i\theta} \mathrm{d}\theta$$ where the integrand of the last integral $$\frac{8R^3e^{3i\theta}-9R^2e^{2i\theta}+6Re^{i\theta}-1} {2R^4e^{4i\theta}-3R^3e^{3i\theta}+3R^2e^{2i\theta}-Re^{i\theta}+1} iRe^{i\theta} \to 4i$$ as $$R \to \infty$$ uniformly for $$\theta \in [-\pi/2,\pi/2]$$. Hence $$\lim_{R \to \infty}\int_{C_R}\frac{p'(z)}{p(z)}\mathrm{d}z=4\pi i.$$ It follows that the total change of argument of $$p(z)$$ on $$C_R$$ from $$-iR$$ to $$iR$$ tends to $$4 \pi i$$ as $$R \to infty$$.

For the change of argument on the imaginary axis, we don't compute the integral directly. Instead, we will try to figure out the change of the argument $$Arg(p(z))$$ at the endpoints of $$I_R$$.

Since $$p(z)$$ has no roots on $$I_R$$, we have that

$$\int_{I_R}\frac{p'(z)}{p(z)}\mathrm{d}z=Arg(p(iR))-Arg(p(-iR)).$$

Since $$p(iR)=2R^4-3R^2+1+i(3R^3-R),$$ we have that $$Arg(p(iR))=\arctan(\frac{3R^3-R}{2R^4-3R^2+1}) \text{ and } Arg(p(-iR))=\arctan(\frac{-3R^3+R}{2R^4-3R^2+1}).$$ Hence the total change of argument of $$p(z)$$ on $$I_R$$ from $$iR$$ to $$-iR$$ equals $$2\arctan(\frac{-3R^3+R}{2R^4-3R^2+1}) \to 0 \text{ as } R \to \infty.$$

In conclusion, the total argument change of $$p(z)$$ on the contour $$T_R$$ tends to $$4\pi i$$ as $$R \to \infty$$, which implies that there are exactly two roots on the right half plane.

Certainly all four roots of $$f(-z)$$ are neither pure real or pure imaginary. Therefore either $$f(z)$$ is stable, $$f(-z)$$ is stable, or the roots of $$f(z)$$ lie one in each quadrant.

$$f(z)$$ is not stable: in the Hurwitz criterion for $$f(z)$$ we have $$D_1=-1<0$$.

$$f(-z)$$ is not stable: in the Hurwitz criterion for $$f(-z)$$ we have $$D_2=0$$.

Therefore the roots of $$f(z)$$ lie one in each quadrant.

• What does stable mean? – lhf Oct 24 '18 at 0:59
• A polynomial is stable if each of its roots has strictly negative real part. – K B Dave Oct 24 '18 at 1:12
• Would not have thought of the Hurwitz criterion. Nicely done. – eyeballfrog Oct 24 '18 at 4:43

I offer a long, tedious, boring solution. I hope there is an elegant proof of this problem, and someone will post it.

Suppose that the roots of $$p(z)=2\,\left(z^4-\frac{3}{2}\,z^3+\frac{3}{2}\,z^2-\frac12\,z+\frac12\right)$$ are $$a\pm b\text{i}$$ and $$c\pm d\text{i}$$ for some $$a,b,c,d\in\mathbb{R}$$ (we already know that $$p(z)$$ has no real roots). We may further assume that $$b$$ and $$d$$ are positive (but this is irrelevant to the proof of this problem). Thus, $$p(z)=2\,\big((z-a)^2+b^2\big)\,\big((z-c)^2+d^2\big)\,,$$ so that \begin{align}p(z)&=2\,\Biggl(z^4-2(a+c)\,z^3+\left(a^2+b^2+c^2+d^2+4ac\right)\,z^2\\&\phantom{aaaaa}-2\Big(\left(a^2+b^2\right)c+\left(c^2+d^2\right)a\Big)\,z+\left(a^2+b^2\right)\left(c^2+d^2\right)\Biggr)\,.\end{align} Consequently, if $$m:=ac$$, $$x:=a^2+b^2$$, and $$y:=c^2+d^2$$, then $$a+c=\frac34\,,\tag{1}$$ $$x+y=\frac{3}{2}-4m\,,\tag{2}$$ $$cx+ay=\frac14\,,\tag{3}$$ and $$xy=\frac12\,.\tag{4}$$

From (2) and (3), we get $$(a-c)x=\frac{3}{2}a-4ma-\frac14\text{ and }(c-a)y=\frac{3}{2}c-4mc-\frac14\,.$$ Multiplying the two results above and using (4), we obtain $$-\frac{1}{2}(a-c)^2=\left(\frac{3}{2}a-4ma-\frac14\right)\left(\frac{3}{2}c-4mc-\frac14\right)\,.$$ From (1), we can see that the previous equation is equivalent to $$-\frac{1}{2}\left(\left(\frac{3}{4}\right)^2-4m\right)=16m^3-12m^2+3m-\frac{7}{32}\,.$$ Hence, $$m$$ is a real root of the cubic polynomial $$q(t):=16t^3-12t^2+t+\frac{1}{16}\,.$$

Note hat $$q'(t)=48t^2-24t+1$$, so $$q'(t)$$ has two roots $$\dfrac{3-\sqrt{6}}{12}>0.045>\dfrac{1}{40}\text{ and }\dfrac{3+\sqrt{6}}{12}<0.455\,.$$ Since $$\lim\limits_{t\to-\infty}\,q(t)=-\infty$$, $$q(0)=\dfrac{1}{16}>0$$, $$q\left(\dfrac12\right)=-\dfrac{7}{16}$$, and $$\lim\limits_{t\to+\infty}\,q(t)=+\infty$$, we conclude that $$q(t)$$ has three real roots $$u<0$$, $$v\in (0.045,0.455)$$, and $$w>\dfrac12$$. (Numerically, $$u\approx -0.04111$$, $$v\approx 0.14768$$, and $$w\approx 0.64343$$.)

We may assume without loss of generality that $$a\geq c$$. Since $$a+c=\dfrac34$$, it follows immediately that $$a>0$$. Since $$0$$ is not a root of $$q(t)$$, $$c\neq 0$$. We shall prove that $$c<0$$ and the claim that $$p(z)$$ has one complex root in each of the four quadrants is established.

Suppose for the sake of contradiction that $$c>0$$, then by the AM-GM Inequality, we have $$\frac{1}{2}=xy\leq \left(\frac{x+y}{2}\right)^2=\left(\frac{\frac32-4m}{2}\right)^2\text{ and }m\leq \left(\frac{a+c}{2}\right)^2=\frac{9}{64}\,.$$ That is, $$m\leq \dfrac{9}{64}$$, and either $$m\geq \frac{3}{8}+\frac1{2\sqrt{2}}\text{ or }m\leq \frac{3}{8}-\frac1{2\sqrt{2}}\,.$$ Therefore, $$m\leq \frac{3}{8}-\frac{1}{2\sqrt{2}}<0.022<\frac{1}{40}\,.$$ However, since $$m$$ is a root of $$q(t)$$ and $$m>0$$, it must equal $$v$$ or $$w$$, but both $$v$$ and $$w$$ are greater than $$\dfrac1{40}$$. This is a contradiction, and so $$m=u<0$$ must hold. This means $$c<0$$. WolframAlpha confirms this: $$a\approx 0.80130$$, $$b\approx 0.79308$$, $$c\approx -0.05130$$, and $$d\approx 0.62509$$.

Let's do a proof by contradiction. I'm going to start from Batominovski's (1-4), $$\begin{eqnarray} a+c &=& \frac{3}{4}\tag{1}\\ x+y &=& \frac{3}{2}\tag{2} -4m\\ cx + ay &=& \frac{1}{4}\tag{3} \\ xy &=& \frac{1}{2}\tag{4} \end{eqnarray}$$ and assume $$m = ac \ge 0$$. Since $$m \ge 0$$, it follows from (2) that $$x + y \le 3/2$$, and from (4) we have $$x + \frac{1}{2x} \le \frac{3}{2}\Longrightarrow2x^2+1\le3x \Longrightarrow(2x - 1)(x-1) \le 0$$ Thus $$1/2 \le x \le 1$$, and from (4) we have $$1/2\le y \le 1$$. Since, $$x,y\ge 1/2$$, from (3) and (1) we have $$\frac{1}{4} = cx + ay \ge\frac{c}{2} + \frac{a}{2} = \frac{a+c}{2} = \frac{3}{8}$$ Since $$1/4 <3/8$$, we have our contradiction. $$m = ac<0$$, and thus $$c < 0 < a$$ and the four roots $$a+bi$$, $$a-bi$$, $$c+di$$, and $$c-di$$ lie in different quadrants.