According to Wikipedia,
A complex random variable $Z$ on the probability space$(\Omega,{\mathcal {F}},P)$ is a function $Z\colon \Omega \rightarrow \mathbb {C} $ such that both its real part $\Re {(Z)}$ and its imaginary part $\Im {(Z)}$ are real random variables on $(\Omega ,{\mathcal {F}},P)$.
Can we define a complex random variable as follows:
A complex random variable $Z$ on the probability space$(\Omega,{\mathcal {F}},P)$ is a function $Z\colon \Omega \rightarrow \mathbb {C} $ such that $\{\omega \in \Omega |Z(\omega )\in S\} \in \mathcal{F}$ for all $S\in\mathcal{G}$ with $\mathcal{G}$ being the Borel $\sigma$-algebra on $\mathbb{C}$.
Are the above two definitions equivalent?
The second definition is missing something after the "such that" - there is no condition stated?
EDIT: I see you fixed this. Yes, the two conditions are equivalent.
In general, a random variable is none other than a measurable function between two measure spaces (a measure space is a set equipped with a $\sigma$-algebra). Your second definition is a special case of this more general definition. To see why it is equivalent to the first definition, we need to understand the relationship between the Borel $\sigma$-algebras on $\mathbb R$ and $\mathbb C$.
We can use a nice property of the Borel $\sigma$-algebra on any topological space $X$: a function $f\colon (\Omega,\mathcal F)\to (X,\mathcal B(X))$ is measurable if and only if $f^{-1}(U)$ is measurable for all open sets $U\subseteq X$. Now since $\mathbb C$ is equipped with the product topology of $\mathbb R\times \mathbb R$, a set $U$ is measurable if and only if $\pi_1(U)$ and $\pi_2(U)$ are measurable, where $\pi_i$ denote the projections onto the coordinates. Thus, it follows that the first condition is equivalent to the second, as a result of the fact that $\mathbb C$ is equal to the product topology of two copies of $\mathbb R$.