Continuous time Fourier transform and Laurent series:
I recall equations (1) and (2), for convenience set $a=0$, and substitute $z=e^{it/T}$. The function $g(t)=f(e^{it/T})$, with $t\in(-\pi T,\pi T)$, is periodic with period $2\pi T$, given by the Laurent series $$g(t)=\sum_{n=-\infty}^\infty c_n e^{int/T},$$ with coefficients $$c_n=\frac{1}{2\pi T}\int_{-\pi T}^{\pi T}g(t)e^{-int/T}\,dt.$$ For $T\gg t$ the sum over $n$ may be approximated by an integral over $\omega=n/T$ with coefficients $C(\omega)=Tc_{n=\omega T}$, giving $$g(t)=\int_{-\infty}^\infty C(\omega) e^{i\omega t}d\omega,$$ $$C(\omega) =\frac{1}{2\pi}\int_{-\infty}^\infty g(t)e^{-i\omega t}\,dt.$$
In this way the Fourier integral can be obtained as the limit of the Laurent series when the periodicity of the function tends to infinity. Notice that the exponent $nt/T$ cannot be set to zero because, even though $t/T\ll 1$, the product $nt/T=\omega t$ need not be small.
Discrete Fourier transform and Laurent series:
We now start from a discrete time signal $x_n$ and construct the Z-transform $$X(z)=\sum_{n=-\infty}^\infty x_{n}z^{-n}=\sum_{n=-\infty}^\infty x_{-n}z^n.$$ This is a Laurent series centered at $a=0$, with inversion formula $$x_{n}=\frac{1}{2\pi i}\oint X(z)z^{n-1}\,dz.$$ For the discrete Fourier transform one has only $N$ distinct values of $x_n$, the set $\{x_0,x_1,x_2,\ldots x_{N-1}\}$. We extend this set periodically by $x_{n+N}=x_n$. The Z-transform is a series of $N$ Dirac delta functions with coefficients $X_k$, $$X(z=e^{2\pi iq/N})=\sum_{k=0}^{N-1}X_k\delta(q-k),\;\;X_k=\sum_{n=0}^{N-1}x_n e^{-2\pi i kn/N}.$$ The delta function converts the integral in the inversion formula into a sum, $$x_n=\frac{1}{N}\sum_{k=0}^{N-1}X_k e^{2\pi ikn/N}.$$