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.
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 $|\Im z|\to\infty$.