$e^{it}$ is not analytic?

Question

Is the following reasoning correct? $f(t)=e^{it}$ with $t \in \mathbb{R}$, is not analytic because $f:\mathbb{R}\rightarrow\mathbb{C}$ so the complex derivative is not defined for an open set in $\mathbb{C}$. (Only for open sets in $\mathbb{R}$.)