How many roots does \[ z^4+3z^2+z+1 \] have in the right half-plane?
Solution
Set \[ p(z)\coloneqq z^4+3z^2+z+1. \]
1. The polynomial \(p\) has no zero on the imaginary axis.
For \(y\in\RR\), \[ p(iy) = y^4-3y^2+1+iy. \] If this were zero, its imaginary part would give \(y=0\), but then its real part would be \(1\). Hence \(p(iy)\neq0\) for every real \(y\).
2. Choose \(R>0\) so large that every zero of \(p\) lies in \(\abs{z}<R\) and\[ 3R^2+R+1<R^4. \]Set\[ A_R\coloneqq R^4-3R^2+1 \]and let\[ \alpha_R\coloneqq\arg(A_R+iR)\in(0,\pi/2). \]Along the diameter of the right half-disk, traversed from \(iR\) to \(-iR\), the change in argument of \(p\) is\[ -2\alpha_R. \]
The displayed inequality implies \(A_R>0\). By step 1, the path \[ y\longmapsto p(iy) = y^4-3y^2+1+iy \] never passes through the origin.
For \(y>0\) it lies in the open upper half-plane, for \(y<0\) it lies in the open lower half-plane, and \(p(0)=1\). Hence the principal argument is continuous along this path from \(y=R\) down to \(y=-R\). At the endpoints, \[ p(iR)=A_R+iR, \qquad p(-iR)=A_R-iR, \] whose principal arguments are \(\alpha_R\) and \(-\alpha_R\), respectively. Thus the change is \(-2\alpha_R\).
3. Along the semicircle\[ z=Re^{i\theta}, \qquad -\frac\pi2\leq\theta\leq\frac\pi2, \]traversed from \(-iR\) to \(iR\), the change in argument of \(p\) is\[ 4\pi+2\alpha_R. \]
On \(\abs{z}=R\), write \[ p(z) = z^4\left( 1+\frac{3z^2+z+1}{z^4} \right). \] The choice of \(R\) gives \[ \left| \frac{3z^2+z+1}{z^4} \right| \leq \frac{3R^2+R+1}{R^4} <1. \] Therefore the factor in parentheses lies in the open disk of radius \(1\) centered at \(1\), so it never vanishes and has a continuous principal argument on the semicircle.
For \(z=Re^{i\theta}\), choose the continuous argument \[ 4\theta+ \arg\left( 1+\frac{3z^2+z+1}{z^4} \right) \] for \(p(z)\). The first term changes from \(-2\pi\) to \(2\pi\), contributing \(4\pi\).
At \(z=-iR\), the second factor is \(p(-iR)/R^4\), whose argument is \(-\alpha_R\); at \(z=iR\) its argument is \(\alpha_R\). Hence the second term contributes \(2\alpha_R\). The total change is therefore \(4\pi+2\alpha_R\).
4. The change in argument of \(p\) around the positively oriented boundary of the right half-disk\[ \{z:\operatorname{Re}z>0,\ \abs{z}<R\} \]is \(4\pi\).
The positive orientation traverses the diameter from \(iR\) to \(-iR\) and then the right semicircle from \(-iR\) to \(iR\). Adding steps 2 and 3 gives \[ -2\alpha_R+(4\pi+2\alpha_R)=4\pi. \]
5. The polynomial \(p\) has exactly two roots in the right half-plane, counted with multiplicity.
The boundary used in step 4 contains no zero of \(p\): step 1 handles the diameter, while step 3 shows that \(p\) is nonzero on the semicircle. By the argument principle, the number of zeros in the right half-disk is \[ \frac{4\pi}{2\pi}=2. \] The radius \(R\) was chosen larger than the modulus of every zero of \(p\), so these are exactly the roots of \(p\) in the entire right half-plane.
6. Q.E.D.
Step 5 gives the required answer: \(2\) roots.