Prove that the equation $az^3-z+b=e^{-z}(z+2)$ has two solutions in the right half-plane $\{z\in\mathbb{C}\,:\,\Re z>0\}$ when $a>0$ and $b>2$. 
Prove that the equation
  $$
az^3-z+b=e^{-z}(z+2)
$$
  has two solutions in the right half-plane $\{z\in\mathbb{C}\,:\,\Re z>0\}$ when $a>0$ and $b>2$.

This is an old qualifying exam question. I'm sure it uses Rouché's theorem somehow, but I can't quite figure it out. From my experience, the way Rouché's theorem works is that you basically create an equation that you know has the right number of zeroes by adding/subtracting terms from your original equation, then compare it to your original. The problem is I can't create an equation which has the number of zeroes I want. The best choice I have is comparing it to $az^3-z+b$, but after I would even show that it has the same number of zeroes as $az^3-z+b-e^{-z}(z+2)$, I would have to somehow justify that $az^3-a+b$ has $2$ zeroes in the right half-plane.
I also considered using the conformal map $z\mapsto\frac{1-z}{1+z}$ and working with the unit disk as the domain which I generally find to be easier to work with (as opposed to using some arbitrarily large rectangle), but it doesn't seem to help.
Any help is greatly appreciated. Thanks in advance.
 A: Comparing to $f(z)=az^3-z+b$ will work for you since $f$ has two roots in the right half-plane, which we can prove as follows. First, this function must have a negative real root since $f(0)=b>0$ and $\lim_{z\to -\infty}f(z)=-\infty$. Using Vieta's formula for the coefficient of $z^2$ (here $0$), given roots $p_1,p_2,p_3$ we must have $p_1+p_2+p_3=0$ so $\Re p_1+\Re p_2+\Re p_3=0$. Vieta's formula also gives $p_1p_2p_3=-\frac{b}{a}<0$. This means that if $p_1<0$, neither $p_2$ nor $p_3$ can be negative real numbers because the negativity of one would contradict the negativity of $p_1p_2p_3$ and the negativity of both would contradict $p_1+p_2+p_3=0$. Thus either $p_1<0<p_2<p_3$ (so we are done), or $p_2$ and $p_3$ have nonzero imaginary part and thus $p_3=\overline{p_2}$. But then $\Re p_1+\Re p_2+\Re p_3=p_1+2\Re p_3=0$, so $\Re p_2=\Re p_3=-\frac{1}{2}p_1>0$, proving the claim.
From there the trick is to use a rectangle whose left side is the imaginary axis and whose dimensions approach $\infty$, and apply Rouché's theorem. For the imaginary axis, use
$$
|f(ki)|^2=|a(ki)^3-ki+b|^2 = |(-ak^3-k)i+b|^2=k^2(ak^2+1)^2+b^2
$$
while
$$
|e^{-ki}(ki+2)|^2=|ki+2|^2=k^2+4<k^2(ak^2+1)^2+4<|f(ki)|^2.
$$
For the right side approaching $\infty$ in real part, use the fact that $|e^{-z}(z+2)|=e^{-\Re z}|z+2|\to 0$, and the bounds for the top and bottom should be reasonable since $f$ grows faster than $|z+2|$ when you consider $|\Im z|\to\infty$.
