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

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.

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.