Solution: We recall that in the cyclic group $\mathbb{Z} _n$, the element $\bar{r} $ is a generator if $\gcd (n,r) = 1$. Thus the set of all generators in $\mathbb{Z} _n$ is given by \[ \mathcal{R} = \left\{ \bar{r} \in \mathbb{Z} _n \mid \gcd (n,r) = 1 \right\} \] The number of elements in the set $\mathcal{R} $ is the Euler-phi function, $\varphi $. Thus, the number of generators in $\mathbb{Z} _{2025}$ is $\varphi (2025)$. Recall that if $\gcd (a,b) = 1$, then $\varphi (a\cdot b) = \varphi (a) \cdot \varphi (b)$. Also, for any prime $p$, \[ \varphi \left( p^k \right) = p^k - p^{k-1}. \] Note that \begin{align*} 2025 = 3^4 \times 5^2 \implies \varphi (2025) & = \varphi \left( 3^4 \right) \cdot \varphi \left( 5^2 \right) \\ & = (3^4 - 3^3) \cdot (5^2 - 5) \\ & = (81 - 27) \cdot (25 - 5) \\ & = 54 \cdot 20 \\ & = 1080. \end{align*}