Using Green's theorem in conjunction with Cauchy-Riemann Equations? 
This is a problem from Stein's Complex Analysis -- I'm not really sure what's happening here.
How precisely does $dz = (dx + idy)$, given that $f(z) = F(x, y) + iG(x, y)$? 
Furthermore, the original integral is split into two integrals, both of which are integrated over $T^0$; what exactly does $T^0$ represent?
Thanks.
 A: If $\gamma(t) = x(t)+iy(t)$ is a parametric representation of $\gamma$, then $\int_{\gamma}f(z)\,dz$ is defined to be a Riemann-Stieltjes integral. If
$\Delta_{k}\gamma$ denotes $\gamma(t_{k+1})-\gamma(t_{k})$ and a similar expression for $\Delta_{k}x$, $\Delta_{k}y$, and $\gamma_{k}^{\star}$ denotes $\gamma(t_{k}^{\star})$, etc., then
$$
       \int_{\gamma}f(\gamma(t))\,d\gamma(t)=\lim_{\|\mathcal{P}\|\rightarrow 0}\sum_{\mathcal{P}}f(\gamma_{k}^{\star})\Delta_{k}\gamma \\
   = \lim_{\|\mathcal{P}\|\rightarrow 0}\sum_{\mathcal{P}}\left[(F(x_{k}^{\star},y_{k}^{\star})+iG(x_{k}^{\star},y_{k}^{\star})\right]
      \left[\Delta x_{k}+i\Delta y_{k}\right] \\
   = \lim_{\|\mathcal{P}\|\rightarrow 0}\left[\sum_{\mathcal{P}}F(x_{k}^{\star},y_{k}^{\star})\Delta x_{k}
     +i\sum_{\mathcal{P}}G(x_{k}^{\star},y_{k}^{\star})\Delta x_{k}+\cdots\right] \\
     = \int_{\gamma}F(x,y)\,dx +i\int_{\gamma}G(x,y)\,dx+i\int_{\gamma}F(x,y)\,dy-\int_{\gamma}G(x,y)\,dy\\
    = \int_{\gamma}Fdx-Gdy + i \int_{\gamma}Gdx +Fdy.
$$
So the notation of the integral appears to expand in an algebraic way because the discrete $\Delta's$ expand in the same way, and the notation is well-paired with the discrete. Now Green's Theorem may be applied, assuming that $F$ and $G$ are continuously differentiable on and in $\gamma$.
A: I'm not sure what your problem is with $dz = dx + idy$ and $f = F + i G$.
Here $T^\circ$ denotes the interior of the triangle whose boundary is $T$. There are typos in your book: in the two first lines of the series of equalities, there should only be $T$s, and no $T^\circ$s.
