$G$ finite group and $\mathbb{C}[G]$ its group ring, show: $K_i = \sum_{a_k\in cl_G(a_i)} a_k $, then $Z(\mathbb{C}[G])=span_\mathbb{C}(K_1,...,K_n)$

Question

Given a finite group $G = \{a_1 ,a_2,...,a_n\}$ and $\mathbb{C}[G]$ the respective group ring, $\mathbb{C}[G] = \{\sum z_ia_i : z_i \in\mathbb{C} , a_i\in G\}$.

Defining $K_i := \sum_{a_k\in cl_G(a_i)} a_k$ , when $cl_G(a_i)$ the conjugacy class, means $K_i =a_1a_ia_1^{-1} + a_2a_ia_2^{-1} +...+a_na_ia_n^{-1}$.

I wish to show that $(K_1 ,K_2 ,...,K_n)$ spans Z($\mathbb{C}[G])$. I had already shown that for $a_i \in Z(\mathbb{C}[G])$, $K_i = \{a_i\}$ thus $a_i = 1\cdot K_i$, as needed.

But I failed to show the other direction, means that for $w\in span_\mathbb{C}(K_1,K_2,...,K_n)$ then $w$ commutes with any element of $\mathbb{C}[G]$, means $w\in Z(\mathbb{C}[G])$.