I am confused on a couple things:
1.) Why is it that an integral of a complex valued function of a complex variable exists if f(z(t)) is piecewise continuous (and/or piecewise continuous on $\mathbb{C}$) and not continuous, like a real?
2.) Why is it that one cannot make use of an antiderivative to evaluate an integral of a function like $1/z$, on a contour of something like $z=2e^{i\theta}$ positively oriented with $-\pi \le \theta \le \pi$? That is, because $F'(z)=1/z$ is undefined at $0$, it is disqualified (I think). But why?
If the function $f(z)$ has an antiderivative $F(z)$ on a domain that includes a contour $\Gamma$ that goes from point $a$ to point $b$, then $\int_\Gamma f(z)\ dz = F(b) - F(a)$. The problem is that in many cases such an antiderivative can't exist. A case in point is your example, $\int_\Gamma \frac{dz}{z}$, where $\Gamma$ is the counterclockwise circle of radius $2$ centred at $0$. You'd like to say that an antiderivative of $f(z) = 1/z$ is $\log(z)$, but there is no version of $\log(z)$ that is continuous on that whole circle. Indeed that must be so, because since $b=a$ (this contour ends in the same place where it starts), you'd have $\int_\Gamma \frac{dz}{z} = F(a) - F(a)=0$. But, parametrizing the contour as $z(t) = 2 e^{it}$, $$ \int_\Gamma \frac{dz}{z} = \int_{-\pi}^{\pi} \frac{2 i e^{it}\ dt}{2e^{it}} = \int_{-\pi}^\pi i\ dt = 2 \pi i $$