Let $\mu=\sum_{i=1}^n \lambda_i\delta_{w_i}$, where $\delta$ is the Dirac measure, and $\lambda_i,w_i\in\mathbb{C}$ and $\lambda_i$ is non-zero. Let $\sigma_k$ be Fourier coefficients of $f$: $$ \sigma_k=\dfrac{1}{2T}\int_{-T/2}^{T/2} f(x)e^{-2\pi i k x}d\mu. $$ Then why do we have the following equality: $$ \sigma(z)=\sum_{k\in\mathbb{N}}\sigma_k\dfrac{z^k}{k!}=\sum_{i=1}^n\lambda_if(w_i)e^{w_iz}? $$
My work:
\begin{align} \sigma(z)&=\dfrac{1}{2T}\sum_{k\in\mathbb{N}}\dfrac{z^k}{k!}\int_{-T/2}^{T/2}f(x)e^{-2i\pi (kx)}d \mu\\ &=\dfrac{1}{2T}\sum_{k\in\mathbb{N}}\dfrac{z^k}{k!}\int_{-T/2}^{T/2}f(x)e^{-2i\pi (kx)}d \left({\sum_{i=1}^{r}\lambda_i\delta_{w_i}} \right)\\ &=\dfrac{1}{2T}\sum_{k\in\mathbb{N}}\dfrac{z^k}{k!}\sum_{i=1}^r\lambda_if(w_i)e^{-2i\pi (kw_i)}\\ &=\dfrac{1}{2T}\sum_{i=1}^{r}\lambda_if(w_i)\sum_{k\in\mathbb{N}}\dfrac{z^k}{k!}e^{-2i\pi kw_i}\\ &\stackrel{?}{=}\sum_{i=1}^{r}\lambda_if(w_i)e^{w_iz} \end{align}
I don't know where the mistake is in my reasoning or am I misunderstanding something? Can you help me please? Thank you very much.