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, $$g(t)\mapsto \int_{-\infty}^\infty c(\nu) e^{i\nu t/T}d\nu,$$ $$c(\nu) \rightarrow \int_{-\infty}^\infty g(t)e^{-i\nu t/T}dt.$$ Note that the exponent $\nu t/T$ cannot be set to zero, because even though $t/T\ll 1$ the product $\nu t/T$ need not be small.
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.