Number of zeros with positive real part 

How many zeros does the polynomial $z^4 + 3z^2 + z + 1$ have in the right half-plane?


This question is from a previous exam and I am not sure how to solve it. The other questions in this section are applications of Rouché's theorem.
I tried to change this to a question about the unit disc, using the map $z \mapsto \frac{z-1}{z+1}$, which (I think) maps $\{z : \mathfrak{Re}(z) > 0\}$ bijectively to the unit disc and turns rational functions into rational functions. In this way I could reduce to counting the zeros of $4z^4 + 2z^3 + 6z^2 - 2z + 6$ on the unit disc. This looks like a standard Rouché's theorem problem, but I don't see how to do it.
Is it possible that there is a better solution?

I have an attempted proof:
First, $f$ has no zero on the imaginary axis: at any such zero $iy$, we would have $$\mathfrak{Im}(f(iy)) = \mathfrak{Im}(y^4 - 3y^2 + iy + 1) = y = 0,$$ and $0$ is clearly not a zero of $f$.
It is also clear that $f$ has no positive real roots, since it is monotone increasing for $z > 0$.
The derivative $f'(z) = 4z^3 + 6z + 1$ is monotone increasing and therefore has a negative real root.
The second derivative $f''(z) = 12z^2 + 6$ has two purely imaginary roots.
By the Gauss-Lucas theorem, all of these lie in the convex hull of the roots of $f$. Therefore, there exist roots with positive real part, and not all roots have positive real part. There must be exactly one pair - 2 roots.
