Using Laurent Series of a function $f(z)$ around a point $a \in \mathbb{C}$ $$f(z) = \sum^{\infty}_{n=-\infty} c_n(z-a)^n \ \ \ \ (1)$$ where $$c_n = \frac{1} {2\pi i}\int\limits_{\gamma}\frac {f(z)} {(z-a)^{n+1}} dz \ \ \ \ (2)$$ where $\gamma$ is a closed curve around $a$.

And choosing $\gamma$ such that $z$ can be parameterized as $z=a+e^{it}\ t \in [-\pi,\pi] $ in order to obtain Fourier Series

$$ f(t) = \sum^{\infty}_{n=-\infty} c_ne^{int} \ \ \ \ (3) $$ where $$ c_n = \frac{1} {2\pi}\int_{-\pi}^{\pi} f(t)e^{-int}dt \ \ \ \ (4) $$

we can "intuitively" prove Fourier Series by just using introductory complex calculus without any advanced mathematical background in Hilbert-Banach Spaces, Functional Analysis etc.

My question is, how can we derive other forms of Fourier Series (also called Fourier Transforms) such as Discrete Fourier Transform (DFT) and Continuous Time Fourier Transform (CTFT) by a similar manner? The problem here is that DFT equations are not closed contour integrals anymore, just discrete finite sums. I could not wrap my head around how Laurent Series may still work in this discrete case. For the CTFT case we have to let integral limits in $(4)$ to go infinity, which corresponds to infinitely many contour integrals in $(2)$. This seems to be divergent, if I am not terribly mistaken?