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.
Best Answer
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.