In the textbook 'Conformal Invariants : Topics in Geometric Function Theory' page 25, there is the following formula (highlighted in yellow): 
I am very stuck on how Ahlfors manages to get formula (2-1), applying greens formula I get that $\gamma$ depends on $\zeta$, i.e. assuming $\Omega$ is a region bounded by finitely many piecewise $C^1$ jordan curves, I get:
$g(\zeta,\infty) = g'(\frac{1}{\zeta},0)$, where $g'(z,0)$ refers to the greens function with pole at $0$ in the region $i(\Omega)$ where $i(z) = \frac{1}{z}$, I can use greens theorem to figure out a formula for $g'(\zeta,0)$, I get for $\zeta \neq 0$ :
$\log (\frac{1}{|\zeta|}) - g'(\zeta,0) = \frac{1}{2 \pi} \int_{\partial i(\Omega)} \log (\frac{1}{|z- \zeta|}) \frac{\partial g'}{\partial n} |dz|$ where the unit normal $n$ is to the right of the tangent vectors to $\partial i(\Omega)$ (the boundary oriented so that $i(\Omega)$ lies to the left), however I don't know how to go from this to Ahlfors formula, and rather than getting the term $\gamma = G(0,0)$ where $G(z,z_0)$ solves the dirichlet problem in $i(\Omega)$ with boundary values $\log|z-z_0|$, I get additional dependence on $\zeta$, i'm not sure what I'm doing wrong.
Thanks in advance for any help.
I have since solved the problem but likely not in the way ahlfors intended, I did not use greens theorem, instead I used that the functions $\log(|\frac{z}{1 - z \zeta}|)$ and $g'(z,0)$ are both harmonic in $i(\Omega) - \{0 , \frac{1}{\zeta} \}$ (in fact on a slightly larger region containing the closure of this region minus $0, \frac{1}{\zeta}$), so calling them $g_1,g_2$, and letting $c_0,c_{\frac{1}{\zeta}}$ denote small circular contours enclosing $0$ and $\frac{1}{\zeta}$ oriented positively, we have that the cycle $\partial i(\Omega) - c_0 - c_{\frac{1}{\zeta}}$ is homologous to zero in a region containing $\partial i(\Omega)$ and $\Omega - \{0 , \frac{1}{\zeta} \}$ but of course omitting $0 , \frac{1}{\zeta}$, then since $g_1 (^{\ast} dg_2) - g_2(^{\ast} dg_1)$ is locally exact in any region where $g_1,g_2$ are harmonic (here if $u$ is harmonic in some region $D$, then $^{\ast} du = -u_y dx + u_x dy$, we can define a single valued harmonic conjugate of $u$ if and only if this differential is exact in $D$ (it is of course always locally exact) ) integrating this differential over this cycle yields zero, and we will eventually get the desired result, then we translate to the unbounded region by means of the mapping $z \rightarrow \frac{1}{z}$ and it seems to all work out.
I would be very grateful if someone could spell out the strategy to approach this with green's theorem since our region is unbounded I wasn't able to immediately apply it (I'm also not very familiar with stokes/green's owing to some lack of basic knowledge of differential forms that I have yet to fill), but I am guessing if we map this unbounded region conformally onto a bounded region by means of $z \rightarrow \frac{1}{z}$. then we could apply green's theorem for multiply connected domains given we choose our functions carefully.