"Hi: I am reading "complex variables" by Ash and Novinger and they prove "cauchy's theorem for triangles early in the book". Unfortunately, there's a step in their proof that I don't follow.
Assuming that $f(z)$ is analytic at $z_{0}$, they first write down the well known relation in their equation (2), namely
$f(z) = f(z_{0}) + (z - z_{0})[f^{\prime}(z_{0}) + \epsilon (z)] , z \in \Omega$
Then, on the next line and based on (2) a previous theorem from earlier in the book, they then write
$\int_{T_{n}} f(z) dz = \int_{T_{n}} (z - z_{0})\epsilon(z) dz, ~~n = 1,2,3 \ldots$ . *
The theorem that they refer to earlier in the book is the following:
Let $z_{1}, z_{2} \in C $ and let $\gamma$ be any path from $z_{1}$ to $z_{2}$, that is $\gamma: [a,b] \rightarrow C$ is any path such that $\gamma(a) = z_{1}$ and $\gamma(b) = z_{2}$. Then, for $n = 0,1,2,3, \ldots $, we have:
$ \int_{\gamma} z^{n} dz = (z_{2}^{n+1} - z_{1}^{n+1})/(n+1). $
Thanks for any help on how they get the equation ending with *.
The earlier theorem implies that
$$\int_{\Gamma} (z - z_0) dz = 0$$
whenever $\Gamma$ is a closed curve. To see this, choose any two points $a$ and $b$ on $\Gamma$, and integrate along $\Gamma$ moving clockwise from $a$ to $b$, and integrate moving counterclockwise from $a$ to $b$. The theorem then allows you to compute the two integrals (and the total integral is the sum of these two), and you should see that they are the same except for a sign.
So upon adding them together, you get $0$. Now just notice that each triangle $T_n$ is a closed curve (in fact, you can explicitly decompose into two curves: Take one side, and then the other two to close it), and the result follows.