Let $V$ be an irreducible $\mathbb CG$ module. We define $Z(G)$ to be the centre of $G$. For $z\in Z(G)$ show that there exists $\lambda_z\in\mathbb C$ such that $z\cdot v=\lambda_z\cdot v$ for all $v \in V$.
I might have done this but I am not sure if my answer is correct.
Recall Schur's Lemma:
Let $V$ and $W$ be irreducible $\mathbb CG$-modules.
i) Suppose $\theta \colon V \to W$ is a $\mathbb CG$-homomorphism. Then either $\theta = 0$ or $\theta$ is an isomorphism.
ii) If $\theta\colon V \to V$ is an isomorphism then there exists $\lambda \in \mathbb C$ such that $\theta(v)=\lambda v$ for all $v\in V$.
So we take a mapping $p\colon V \to V$ where $v\mapsto zv$. This is a $\mathbb CG$ homomorphism. We use Schur's Lemma, using part (i); Since $p$ is a $\mathbb CG$ homomorphism we can say that $p=0$ or $p$ is an isomorphism. If we take $p,g\in G$, then we have $gp=pg$ for all $g\in G$, so $p$ is in $Z(G)$, the centre of $G$. Hence $Z(G)$ is not empty. So $p$ is an isomorphism.
Next we use (ii). There exists $\lambda_z$ in $\mathbb C$ such that $p(V)=\lambda_z$ and since $p(v)=zv$, then we have $zv=\lambda_z$ as required.