The comment box is a bit too brief for this discussion, let me use the answer box. 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.