$\mathbb{E}(X^2) = 1$ implies the characteristic function is equal to $1$. [closed]

We let $X$ be a real random variable. And the characteristic function for $X$ is the function $\varphi: \mathbb{R} \rightarrow \mathbb{C}$ defined by $\varphi(u) = E(e^{iuX})$ for all $u \in \mathbb{R}$. We have $\mathbb{E}(X^2) = 1$. We want to show $\varphi(u) = 1$ for all $u \in \mathbb{R}$.
Since $\mathbb{E}(X^2) = 1$, we then have $\mathbb{P}(X = 0) = 1$. Then we consider $\varphi(u) = \mathbb{E}(e^{iuX})$. Since $\mathbb{P}(X = 0) = 1$, we have $\mathbb{P}(e^{iuX} = e^{iu0}) = 1$. But I'm not sure if I conclude from here that $\varphi(u) = 1$ for all $u \in \mathbb{R}$. Any help is appreciated! Thanks!