Q:

Order of the group generated by $\langle i\rangle$ in $\mathbb{C}^{\ast}$

Let $\mathbb C^{\ast} := \mathbb C \setminus \{0\}$.
So $\mathbb C^{\ast}$ is a group closed under multiplication.
Consider $\langle i\rangle$ which generates $i$.
So,

$i$, $i^2$, $i^3$, $i^4$, $...$, $i^n$ are generated.

Which is the equivalent to:

$i$, $-1$, $-i$, $1$, $...$, $i^n$ are generated.
$1$ is the multiplicative identity.

My guess is that since $\langle i\rangle$ generates identity, that means we can generate the entire group. So the order is infinite.

A:

First, let me take a moment to address your confusions.
That would be the right reasoning for an ideal $I$ in a ring $R$; if $1\in I$, then $I = R$, because an ideal must satisfy $rx\in I$ for any $r\in R$ if $x\in I$. And if $1\in I$, then $r\cdot 1 = r\in I$ for any $r\in R$, implying that $R = I$.
However, we are talking about groups (not rings), and groups have only one identity, as they have only one binary operation (rings have two, addition and multiplication). Any subgroup of $G$ generated by any set of elements must contain the (only) identity of $G$ (as any subgroup contains the identity), and containing the identity does not mean it contains the entire group. For example, consider any group $G$, and the subgroup generated by the identity $e$. Well, $\{e\}$ is already a subgroup, so the subgroup generated by $e$ is just $\{e\}$ ($\langle e\rangle = \{e\}$)!
A group generated by a single element $g$ is necessarily cyclic, since
$$
\langle g\rangle = \{g^n\mid n\in\Bbb Z\},
$$
and this admits a surjective map from $\Bbb Z$ given by
\begin{align*}
\Bbb Z&\to \langle g\rangle\\
n&\mapsto g^n.
\end{align*}
Thus, $\langle g\rangle$ is a quotient of $\Bbb Z$, and is hence cyclic of some order. (Note that this implies that $\langle i\rangle$ cannot be isomorphic to $\Bbb C^\times$, as $\Bbb C^\times$ is not cyclic.)
Back to the problem at hand. We want to determine what $\langle i\rangle\subseteq\Bbb C^\times$ is. Now, we observe that $i^4 = 1$, which is the identity. So, $i^0 = 1$, $i^1 = i$, $i^2 = -1$, $i^3 = -i$, $i^4 = 1$, $i^5 = i$, and so on (similarly for negative powers of $i$). So the set $$\langle i\rangle = \{i^n\mid n\in\Bbb Z\} = \{\dots, i^{-2}, i^{-1}, i^0, i^1, i^2,i^3, i^4,\dots\} = \{\dots, -1,-i,1, i, -1,-i, 1,\dots\}$$ is really simply the set $\{1, i, i^2, i^3\}$. This tells us that $\#\langle i\rangle = 4$ (the order of $i$), and in particular, that $\langle i\rangle $ is isomorphic to the cyclic group of order $4$.

