Elementary question on complex analysis: What domains do the following $z$ take? 
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*Let $C$ be a smooth curve , $a \leq t \leq b$ and suppose $f$ is continuous at all points on $C$ then the integral of $f(z)$ along $C$ is defined by $$\int_{C}f(z) dz = \int_a^{b}f(z(t))z^{'}(t)dt$$
For this Theorem above, we see that the domain of $f$ is along the curve $C$, so the values of $z$ should all lie on the curve $C$, right?


*If $f$ is continuous on a smooth curve $C$ of length $L$ and if $|f(z)| \leq M$ for all $z$ on $C$, then $$|\int_{C}f(z)dz|\leq ML$$
For this ML inequality theorem, i can say that $z$ takes value on the curve $C$ as well if i am not wrong.


*Suppose that a function $f$ is analytic at all points within and on a simple closed contour $C$ in $D$, then $$\int_{C}f(z)dz = 0$$
Above it the Cauchy Goursat, now for this the domain of $f$ should be on and within $C$ hence $z$ takes values on and within $C$ as well? Different from the first 2?


*Same goes for the Cauchy integral formula where the formula is $$f(z_0) = \dfrac{1}{2\pi i}\int_{C}\dfrac{f(z)}{z-z_0}dz$$
Now i think that values of $z$ for the integrand should be also within and on the curve $C$
Now it should be very elementary to not get confused over a simple domain, however let me show you a question i did and my solution presented. Notice the part i circled on my solution, the question is
Let $f(z)$ be an analytic function on a region $D$ and let $a,b \in D$.
a) Show that $$f(b)-f(a) = \dfrac{1}{2\pi i}\int_{C}\dfrac{f(\zeta)}{\zeta - b}d\zeta - \dfrac{1}{2\pi i}\int_{C}\dfrac{f(\zeta)}{\zeta - a}d\zeta$$
where $C$ is a simple closed circle of radius R that encloses both $a$ and $b$.
b) Deduce that if $f(z)$ is a bounded entire function then $f(z)$ must be a constant. (Another proof of Liouville)
Main concern is " domain of $f$" or rather the values that $z$ take!  See my working which says $|z| = R$,which means $z$ is lying on the circle $C$, however before ML formula came in isn't my values of $z$ on and within C? How come now the values of $z$ just strictly lie on $C$?


In a contour integral $\int_C f\ dz$, z lies on the curve $C$ just like $x$ lies in $[a,b]$ in the Riemann integral $\int_a^b f\ dx$. In fact, the definition of the contour integral is a generalized version of the Riemann integral. 
You should not confuse the domain of $f$ with the integration path $C$.
A nice way to think of the Cauchy formula is that given that $f$ is holomorphic on a disk $D$ we can find the value $f(z_0),\ z_0 \in D$ using only the values of $f$ on $\partial D$.
In your proof you should emphasize the fact that you can assume $R\to \infty$ only because $f$ in entire and therefore the Cauchy formula holds for a circle $C$ of arbitrarily large radius $R$.
