Rouche's theorem I want to show that the equation $az^{3}-z+b=e^{-z}(z+2)$, where $a>0$ and $b>2$, has two roots in the right half-plane $\mathrm{Re}\;z\geq 0$. This is a problem in using Rouche's theorem but I am unable to get the right estimates.
I tried taking a semicircular contour in the right half-plane with its arc going from $-iR$ to $iR$ and then going down the imaginary axis. To apply Rouche's theorem, I need holomorphic functions $f,g$ such that $|g(z)|<|f(z)|$ for $z$ on the semicircle. I took $f(z)=az^{3}-z+b$ and $g(z)=e^{-z}(z+2)$. Things seem to be fine on the arc but I ran into problems on the imaginary axis.

I dont think there is a problem on the imaginary axis. You want to find a bound for   $$|e^{-z} (z+2)|$$ 
But $|e^{-z}|=e^{-x}=1$ since $x=0$ on the imaginary axis. Letting $z=iy$ we have
$$|g(z)|^2=y^2+4$$ and
$$|f(z)|^2=y^2(a y^2+1)^2+b^2$$
It is clear now that $|g(z)|<|f(z)|$.
