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

Question

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?

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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.

Answer 1

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.

Answer 2

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 $d\approx 0.62509$.

Answer 3

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.