Solution: We need to find all idempotent elements in $\mathbb{Z} _5[\iota ]$. Recall that $a$ is said to be idempotent if $a^2 = a$.

Let $I(R)$ denote the set of all idempotent elements in $R$. Then, \begin{equation}\label{eq:17July2024-1} I\left( \mathbb{Z} _5[\iota ] \right) = \left\{ a + b \iota \in \mathbb{Z} _5 [\iota ] : (a + b \iota )^2 = a + b \iota \right\}. \end{equation} Note that if $a,b\in \mathbb{Z} _5$ with \begin{align*} (a + b \iota )^2 = (a + b \iota ) & \implies a^2 - b^2 + 2ab \iota = a + b \iota \\ & \implies (a^2 + b^2 -a) + (2a - 1)b \equiv 0 \text{ mod } 5\\ & \implies a^2 - b^2 - a \equiv 0 \text{ mod } 5 \\ & \kern 1cm \text{ and } (2a - 1)b \equiv 0 \text{ mod } 5. \end{align*} If $(2a - 1) b \equiv 0 \text{ mod } 5$, then as $\mathbb{Z} _5$ is an integral domain, \begin{align*} 2a - 1 = 0 \text{ or } b = 0 \implies a = 3 \text{ or } b = 0 \text{ in } \mathbb{Z} _5 . \end{align*} If $a \equiv 3 \text{ mod } 5$, then \begin{align*} a^2 - b^2 -a =0 & \implies 9 - b^2 - 3 = 0 \\ & \implies 6 - b^2 = 0 \\ & \implies b^2 = 1 \\ & \implies b = 1, 4. \end{align*} If $b \equiv 0 \text{ mod } 5$, then \begin{align*} a^2 - b^2 - a = 0 \implies a ^2 = a \implies a = 1. \end{align*} Thus, \[ I\left( \mathbb{Z} _5 [\iota ] \right) = \left\{ 1, 3 + \iota , 3 + 4 \iota \right\} . \]