Prove $P(z)=z^4+2z^3+3z^2+z+2$ has exactly two zeros in the right half plane NOTE: The answer found here is not what I'm looking for.
The question is:
Prove $P(z)=z^4+2z^3+3z^2+z+2$ has exactly two zeros in the right half plane. [Hint: Write $P(iy)=(y^2-2)(y^2-1)+iy(1-2y^2)$, and show that $\lim_{R\to\infty}arg\{P(iy)\}\Biggr|_{-R}^{R}=0$.]
I need to prove this (preferably using Rouche's theorem. Any other method involving the hint may also be accepted). Now, this can be shown to be true by taking two functions $f(z)=(z-1)^2(z+2)^2$ and $h(z)=z^4+2z^3-3z^2-4z+4$ which are analytic on and inside a closed semi-circular contour (radius $R$) encapsulating the right half-plane, and by showing that $|f(z)|>|h(z)|$ on the boundary of it. Then by Rouche's theorem, it is easy to verify that $P(z)$ has only 2 zeros (same as $f(z)$) in the right half-plane.
What I'm looking for is a solution involving the Hint! I know the Hint says that the imaginary axis of the domain gets mapped to the right half-plane (more precisely to the positive real numbers as $R\to\infty$). But I'm not sure how to use it in getting a solution.
Thanks in advance.
 A: For $R>0$, let ${\cal C}_R$ be the closed curve oriented anti-clockwise and consisting of the segment $S_R=[iR,-iR]$ and the half circle $C_R=\{Re^{it}: t\in [-\pi/2,\pi/2]\}$. It suffices to show that when $R$ goes to infinity, the index $\oint_{{\cal C}_R} \frac{P'(z)}{P(z)} \frac{dz}{2\pi i}$ is well-defined and equals two, or equivalently, writing $P(z)=|P(z)| e^{i\phi}$ that the angular increase $\oint d\phi = 4\pi$. Note that as long as $P(z)\neq 0$,  $d\phi$ is well-defined as a differential form although $\phi$ is only defined mod $2\pi {\Bbb Z}$.
For $R$ large, the dominant term is $z^4$ and $(\pm iR)^4 = R^4 >0$ points in the direction of the positive real axis. Going along the half-circle we have $$\int_{C_R} \frac{P'(z)}{P(z)} \frac{dz}{2\pi i} = 
\int_{C_R} \frac{4}{z} \frac{dz}{2\pi i}+ O(\frac{1}{R}) = 
    \int_{-\pi/2}^{\pi/2} {4} \frac{dt}{2\pi} +O(\frac{1}{R})=
   2 + O(\frac{1}{R}) $$
Regarding the segment $S_R$, the hint implies that $P(iy)$ never points in the direction of the negative real axis. In order to see this, note that for $P(iy)$ to be real the imaginary part must be zero so that $y=0$ or $y^2=1/2$ but in any of these cases the real part is strictly positive (incidentally showing that $P(z)$ never vanishes). The angle thus takes values in $(-\pi,\pi)$ and from the above we infer that $\int_{S_R} d\phi = O(\frac{1}{R})$. Summing and letting $R\to +\infty$ we obtain the wanted result.
A: Let
$$Q(y) = P(iy) = (y^2-1)(y^2-2)+iy(1-2y^2).$$
Since fourth-order polynomial $\;P(z)\;$ has not roots with $\;\Re z=0,\;$ then it suffices to prove that $\;Q(y)\;$ has exactly two roots in the lower half-plane or in the upper one. Let us prove this for the upper half-plane.
Let
$$U(y)=(y^2-1)(y^2-2),\quad V(y)=y(1-2y^2),$$
then
$$\lim\limits_{R\to\pm\infty}\arg Q(y) = \lim\limits_{R\to\pm\infty}\arctan\dfrac{V(y)}{U(y)} = \lim\limits_{R\to\pm\infty}\arctan(\pm 0) = 0.$$
Since $\;Q(-y)=Q(y),\;$ then $\;\lim\limits_{R\to \infty}\arg Q(y)\bigg|_{-R++0i}^{R+0i} = 0.\;$
In accordance with the Rouche's theorem, the functions $\;Q(y) = U(y)+iV(y)\;$ and $\;Q_2(y) = U(y)+2iV(y)\;$ have the same quantity of roots in the upper half-plane.
At the same time,
$$\arg Q_2(y) = \arctan\dfrac{2y-4y^3}{y^4-3y+2} = \arctan \dfrac{\dfrac2{y^3}-\dfrac3y -\dfrac1y}{1+\left(\dfrac2{y^3}-\dfrac3y\right)\dfrac1y}$$
$$= \arctan\dfrac{2-3y^2}{y^3}-\arctan\dfrac1y
 = \arg \left(y^3+i(2-3y^2)\right)+\arg(y-i)$$
$$ = \arg \left(i((iy)^3 +3(iy)^2+2)\right)+\arg(y-i),$$
wherein, in accordance with the Besou theorem, the polynomial
$$t^3+3t^2+2=(t+r)(t^2-(r-3)t+r(r-3))$$
has one real root $\;t=-r,\; r>3\;$ and two complex roots with $\;\Re t>0.\;$
Since in the upper half-plane $\;\Re(iy)<0,\;$ then $\;Q_2(y)\;$ has in sum exactly two roots in the upper half-plane.
Proved!
