Why do we consider here the torus as a subset of the complex numbers?

I'm solving the following exercise:

Show that the n-dimensional Torus $\mathbb{T}^n\subset \mathbb{C}^n$ with i'ts subspace topology is a topological manifold of dimension $n$.

My question is not about how to solve the exercise, I'm only wondering why we take $$\mathbb{T}^n\subset \mathbb{C}^n$$ Wouldn't it work if we take $$\mathbb{T}^n\subset \mathbb{R}^n$$ Or is there another reason?

Thanks for your help.

We usually define $\mathbb{T}^n$ as $\underbrace{S^1 \times S^1 \times \cdots \times S^1}_{n \text{ times}}$. Since $S^1$ is a (topological) subspace of $\mathbb{C}$, the $n$-fold product $\mathbb{T}^n$ is naturally a subspace of $\mathbb{C}^n$, and not $\mathbb{R}^n$. You could, however, identify $\mathbb{C}^n$ with $\mathbb{R}^{2n}$ in the usual way and say that $\mathbb{T}^n \subset \mathbb{R}^{2n}$. Note that requires doubling the dimension/superscript on the right.

In fact, it is a good exercise to prove this cannot possibly be right in low dimensions. For example, $\mathbb{T}^1 = S^1$ cannot embed as a subspace of $\mathbb{R}$. This is a straightforward connectedness argument. Likewise, you can prove that the "standard" torus $\mathbb{T}^2 = S^1 \times S^1$ cannot embed as a subspace of $\mathbb{R}^2$, but it can be realized in $\mathbb{R}^3$, which is a dimension better than the above-described $\mathbb{C}^2 \cong \mathbb{R}^4$.