Find the number of zeroes of the polynomial $h(z)=z^{5} + 5z^{3} + 2z^{2} + 4z + 1$ in the right half-plane. Question: Find the number of zeroes of the polynomial $h(z)=z^{5}+5z^{3}+2z^{2}+4z+1$ in the right half-plane.
Comments: There may be a number of ways to arrive at a solution to this problem, but it would be instructive for me to know if anyone can solve it using Rouché's theorem (or the principle of the argument if that proves impossible). My idea is to count the zeros on an area bounded by the imaginary axis and a half-circle $C$ on the right half-plane with center in $z=0$ and of radius $R$ where $R$ is a large enough number to contain all the zeros in the right-half plane. To do so using Rouché, I would need to see if there is a $g(z)$ which has an absolute value larger than $|f(z)-g(z)|$ on the boundary (that is, on the imaginary axis and on $C$). I am unsure however which $g(z)$ to choose. 
This is a problem from an old complex analysis exam, it is similar to this problem which I posted earlier. All input appreciated.
 A: Choose $g(z) = z^5 + 1$. The zeros of $g$ are $e^{k\pi i/5},\; k \in \{1,3,5,7,9\}$, two of which lie in the right half plane.
Then a slightly generalised version of Rouché's theorem tells you that $h$ also has two zeros in the right half plane.
The standard formulation of Rouché's theorem demands that $\lvert h-g\rvert < \lvert h\rvert$ (or $\lvert h-g\rvert < \lvert g\rvert$) on the boundary $\partial V$ of the region, but looking at the proof, whose main part is the observation that the number of zeros of $f_\lambda = g + \lambda(h-g)$ in $V$,
$$N_\lambda = \frac{1}{2\pi i}\int_{\partial V} \frac{g'(z) + \lambda\bigl(h'(z) - g'(z)\bigr)}{g(z) + \lambda\bigl(h(z) - g(z)\bigr)}\, dz$$
is independent of $\lambda \in [0,\,1]$, since no $f_\lambda$ has a zero on $\partial V$.
The role of the condition $\lvert h-g\rvert < \lvert g\rvert (\text{ or } \lvert h\rvert)$ on $\partial V$ is solely to guarantee that $g + \lambda(h-g)$ also has no zeros on $\partial V$. If we can determine that in a different way, the conclusion still holds.
For a large enough semicircle in the right half plane, we have $\lvert 5z^3 + 2z^2 + 4z\rvert < \lvert z^5 + 1\rvert$ since the RHS is a polynomial of higher degree than the LHS.
On the imaginary axis, although $\lvert g-h\rvert > \lvert g\rvert$ on some parts, expanding shows
$$g(it) + \lambda\bigl(h(it) - g(it)\bigr) = 1 + it^5 + \lambda(-5it^3 - 2t^2 + 4it) = (1-2\lambda t^2) + it(t^4 - 5\lambda t^2 + 4\lambda),$$
and the real part vanishes only for $\lambda = \dfrac{1}{2t^2}$ (with $t^2 \geqslant \frac12$, since $\lambda \leqslant 1$), when the imaginary part is
$$t\left(t^4 - \frac{5}{2} + \frac{2}{t^2}\right) = \frac1t\left(t^6 - \frac{5}{2}t^2 + 2\right),$$
and the polynomial $x^3 - \frac52x + 2$ has no positive real roots, so we conclude $g(z) + \lambda(h(z)-g(z)) \neq 0$ on the imaginary axis.
