 $g(w) \sim w$, so the integrand goes like $1/w^3$ and the contour integral vanishes. 
But we also know that this integral is equal to the integrals we carry out on contours around all the singularities of the integrand. Since these are the branch cut and the point $z$, from (\ref{cauchy}) we find
\begin{align}
    R(z) = -\oint_{C_b} \frac{dw}{2\pi i} \frac{R(w)}{w-z}\frac{g(z)}{g(w)},
\end{align}
with $C_b$ chosen to surround the branch cut and exclude $z$. In particular, we choose
\begin{align}
    R(z) = \frac{g(z)}{2\pi i} \left(\int_{-a}^a dh \frac{R(h+i\epsilon)}{h+i\epsilon-z}\frac{1}{g(h+i\epsilon)}-\int_{-a}^a dh \frac{R(h-i\epsilon)}{h-i\epsilon-z}\frac{1}{g(h-i\epsilon)} \right).
\end{align}
The semicircular boundaries of the contour go to 0 as $\epsilon\to0$. Notice that 
\begin{equation}
    g(h\pm i\epsilon) = \pm i \sqrt{a^2-h^2},
\end{equation} 
which leads to
\begin{align}
    R(z) = \frac{-g(z)}{2\pi} \int_{-a}^a dh \frac{R_+(h) + R_-(h)}{(h-z)\sqrt{a^2 - h^2}}.
\end{align}
Here we can use the Sokhotski–Plemelj theorem and our original integral equation \eqref{eq:integralequation} to write this as
\begin{align*}
    R(z) &= -\frac{\lambda A g(z)}{\pi\chi}\int_{-a}^a dh \frac{h}{(h-z)\sqrt{a^2-h^2}} \\
    &= -\frac{\lambda A g(z)}{\pi\chi} \left( \pi - \frac{\pi}{\sqrt{1 - \frac{a^2}{z^2}}} \right) \\
    &= -\frac{\lambda A}{\chi}\left(z\sqrt{1 - \frac{a^2}{z^2}} -z\right)
\end{align*}
This means
\begin{align}
    R(h \pm i \epsilon) = -\frac{\lambda A}{\chi}\left(\pm i \sqrt{a^2 - h^2} - h \right),
\end{align}
which directly leads to $\rho(h)$. 

\subsection{Strong coupling}

Now we want to find the resolvent defined in (\ref{Rtilde}),
\begin{align}
    \tilde{R}(z) = \int_{(-a,-b)\cup (b,a)} dh \frac{\tilde{\rho} (h)}{z - h}.
\end{align}
 With the choice of $g(z)$ defined in \eqref{gz} we have
\begin{align}
    \oint_{C_\infty} \frac{dw}{2 \pi i} \frac{\tilde{R}(w)}{w-z} \frac{g(z)}{g(w)}=0,
\end{align}
where $C_{\infty}$ is a contour at infinity. Like before, this integral must be equal to the integrals around all singularities of the integrand, so
\begin{align}
    \tilde{R}(z) = -\oint_{C_b} \frac{dw}{2 \pi i} \frac{\tilde{R}(w)}{w-z} \frac{g(z)}{g(w)},
\end{align}
where $C_b$ is a contour that encircles both branch cuts of $g(w)$ counterclockwise.
If we choose this contour to be infinitesimally close to each branch, there is no contribution from the endpoints of the branch cuts and we have
\begin{align}
\begin{split}
    \tilde{R}(z) = \frac{g(z)}{2\pi i} &\left( \int_{-a}^{-b} dh \frac{\tilde{R}(h+i\epsilon)}{(h+i\epsilon-z) g(h+i\epsilon)} -
    \int_{-a}^{-b} dh \frac{\tilde{R}(h-i\epsilon)}{(h-i\epsilon-z)g(h-i\epsilon)} \right.
    \\ &\left. +\int_{b}^{a} dh \frac{\tilde{R}(h+i\epsilon)}{(h+i\epsilon-z) g(h+i\epsilon)} -
    \int_{b}^{a} dh \frac{\tilde{R}(h-i\epsilon)}{(h-i\epsilon-z)g(h-i\epsilon)} \right).
    \end{split}
\end{align}
Now we use the fact that in the region $b < |h| < a$, $g(h-i\epsilon) = -g(h+i \epsilon)$.
This flips the relative sign of the upper and lower branch, so now the values of $R$ above and below the cut are added rather than subtracted:
\begin{align}
\tilde{R}(z) = \frac{g(z)}{2\pi i} \int_{(-a,-b) \cup (b,a)} dh \frac{\tilde{R}(h+i\epsilon) + \tilde{R}(h - i \epsilon)}{(h-z) g(h + i \epsilon)}.
\end{align}
Applying the Sokhotski-Plemelj formula and equation \eqref{strong}, we can eliminate $\tilde R$ from the right-hand side.
We can then reverse the previous process to go back to contour integrals around the branch cuts:
\begin{align}
     \tilde{R}(z) &= \frac{g(z)}{2\pi i} \int_{(-a,-b) \cup (b,a)} dh \frac{2\frac{\lambda A}{\chi} h +2\log\frac{h-b}{h+b}}{(h-z) g(h + i \epsilon)} \\
     &= \frac{g(z)}{2\pi i} \left( \int_{(-a,-b) \cup (b,a)} dh \frac{\frac{\lambda A}{\chi} h + \log\frac{h-b}{h+b}}{(h-z) g(h + i \epsilon)} -  \int_{(-a,-b) \cup (b,a)} dh \frac{\frac{\lambda A}{\chi} h + \log\frac{h-b}{h+b}}{(h-z) g(h - i \epsilon)} \right).
\end{align}
As the numerator has no branch cuts in the region of integration, this is equivalent to the contour integral around each branch cut,
\begin{align}
    \tilde{R}(z) = \frac{-g(z)}{2\pi i} \oint_{C_b} dw\frac{\frac{\lambda A}{\chi}w +\log\frac{w-b}{w+b}}{(w-z)g(w)}.
\end{align}
Now we can consider
\begin{align}
    \oint_{C_\infty} dw\frac{\frac{\lambda A}{\chi}w +\log\frac{w-b}{w+b}}{(w-z)g(w)},
\end{align}
which is zero, but also equal to the sum of contour integrals around all singularities of the integrand. 
So we can again deform the contour:
\begin{align}
    \tilde{R}(z) = \frac{g(z)}{2\pi i} \left(\oint_{C_z} + \oint_{C_l}\right)dw \frac{\frac{\lambda A}{\chi}w +\log\frac{w-b}{w+b}}{(w-z)g(w)},
\end{align}
where $C_l$ is a contour encirculing the branch cut of the logarithm counterclockwise. 
The $C_z$ integral is straightforward thanks to the residue theorem, and the $C_l$ integral is also simple because the jump of the logarithm across the branch cut is constant,
\begin{align}
\tilde{R}(z) &= \frac{\lambda A}{\chi}z + \log\frac{z-b}{z+b} -g(z)\int_{-b}^b \frac{dh}{(h-z)g(h)}.
\end{align}
We can evaluate the integral over $h$ by multiplying the numerator and denominator by $(h+z)$, using parity arguments and a change of variables, to arrive at