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.

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.
