Counting roots of a polynomial in the right half-plane

Let \[ p(z)=z^4+8z^3+3z^2+8z+3. \] On the imaginary axis, \[ p(iy)=\bigl(y^4-3y^2+3\bigr)+8iy(1-y^2). \] The real part is always positive, because with \(t=y^2\), \[ t^2-3t+3=\left(t-\frac32\right)^2+\frac34>0. \] Hence \(p\) has no zeros on the imaginary axis, and as that axis is traversed from \(iR\) down to \(-iR\), a continuous choice of \(\arg p(iy)\) stays in \((-\pi/2,\pi/2)\) and has total change tending to \(0\) as \(R\to\infty\).

Now close the contour by the right semicircle \(z=Re^{i\theta}\), \(-\pi/2\le\theta\le\pi/2\). Uniformly on this arc, \[ \frac{p(z)}{z^4}=1+O(R^{-1}), \] so for large \(R\) the change of argument of \(p\) along the semicircle equals the change for \(z^4\), namely \[ 4\left(\frac\pi2-\left(-\frac\pi2\right)\right)=4\pi. \] Thus the total change of argument around the positively oriented boundary of the right half-disk tends to \(4\pi\). By the argument principle, the number of zeros in the right half-plane is \[ \frac{4\pi}{2\pi}=\boxed{2}. \]