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.
Best Answer
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.