May I know how do we take the contour integral of a constant along a closed triangular region (as defined in Goursat's Theorem)? In other words what does $$\oint_{T} d\zeta$$ equal to?
Also what happens if we suppose we have a line segment L joining points $z_0$ and $z_1$ on the complex plane, and we take the contour integral $$\int_{L} d\zeta$$? Please explain without using assuming that Cauchy-Goursat Theorem holds as I am trying to use the above to prove this theorem. Thank you.
The definition of the line integral is the following :
if you have $f:\Bbb C \to \Bbb C$ and $\gamma(t):[\alpha,\beta]\to \Bbb C$ then the line integral of $f $ over the curve $\gamma$ is the following
$\int_{\gamma} fd\zeta= \int_{t=\alpha}^{\beta} f(\gamma(t))\gamma'(t)dt$ , especially if $F$ is holomorphic and $F'=f$ then
$\int_{\gamma} fd\zeta = \int_{t=\alpha}^{\beta}\frac{\partial}{\partial t}F(\gamma(t))dt=F(\gamma(\beta))-F(\gamma(\alpha)).$
So for example if the curve is closed which happens to the triangle $T$ you have that $\int_{T} d\zeta =0$.
For the line $L=\{(1-t)z_0 +tz_1: t \in [0,1]\}$ you have $\gamma(t)=(1-t)z_0+tz_1$ and $\gamma'(t)=z_1-z_0$.So for the line integral $\int_{L}d\zeta$ you get as $f:\Bbb C \to \Bbb C$ the function $f(u+iv)=1$ then the function $F(z)=z$ is holomorphic and $F'(z)=f(z)$.
So you can write
$\int_{L} d\zeta=\int_{t=0}^{1} f(\gamma(t))\gamma'(t)dt=\int_{t=0}^{1}\frac{\partial}{\partial t}F(\gamma(t))=F(\gamma(1))-F(\gamma(0))=\gamma(1)-\gamma(0)=z_1-z_0.$
I hope that this one helps!