Let $\vec{a},\vec{b}\in \mathbb{C}^n$ be nonzero, where $\vec{a} = (a_1,...,a_n)$ and $\vec{b} = (b_1,...,b_n)$. As a vector space over $\mathbb{R}$, the space $\mathbb{C}^n$ is isomorphic to $\mathbb{R}^{2n}$. That is, for $\vec{a}$ and $\vec{b}$ there corresponds vectors $\vec{x},\vec{y}\in\mathbb{R}^{2n}$ (respectively) such that $ \vec{x} = \begin{pmatrix} \text{Re}\,(a_1) \\ \text{Im}\,(a_1) \\ \text{Re}\,(a_2) \\ \text{Im}\,(a_2) \\ \vdots \ \\ \text{Re}\,(a_n) \\ \text{Im}\,(a_n) \end{pmatrix} \qquad \text{and} \qquad \vec{y} = \begin{pmatrix} \text{Re}\,(b_1) \\ \text{Im}\,(b_1) \\ \text{Re}\,(b_2) \\ \text{Im}\,(b_2) \\ \vdots \ \\ \text{Re}\,(b_n) \\ \text{Im}\,(b_n) \end{pmatrix} \ . $
Recall that $||\,\vec{x}+\vec{y}\,||^2 = ||\, \vec{x}\, ||^2 + ||\,\vec{y}\,||^2+2\,\vec{x}\cdot\vec{y}$ and $ \cos\theta = \frac{\vec{x}\cdot\vec{y}}{||\,\vec{x}\,||\,||\,\vec{y}\,||} \ , $ where $\theta$ is the angle between $\vec{x}$ and $\vec{y}$ (and also the angle between $\vec{a}$ and $\vec{b}$).
$\quad$ We will now show that $\vec{x}\cdot\vec{y} = \text{Re}\,(\vec{a}\cdot\vec{b})$. It is easy to show that $ ||\,\vec{a}+\vec{b}\,||^2 = ||\,\vec{a}\,||^2+||\,\vec{b}\,||^2 + \vec{a}\cdot\vec{b} + \overline{\vec{a}\cdot\vec{b}} $ and $ ||\,\vec{x}+\vec{y}\,||^2 = ||\,\vec{x}\,||^2+||\,\vec{y}\,||^2+2\,\vec{x}\cdot\vec{y}. $ It is also easily show that $||\,\vec{x}\,|| = ||\,\vec{a}\,||$ and $||\,\vec{y}\,||=||\,\vec{b}\,||$. Consequently, $||\,\vec{x}+\vec{y}\,|| = ||\,\vec{a}+\vec{b}\,||$. Therefore, $||\,\vec{a}+\vec{b}\,||^2 = ||\,\vec{a}\,||^2+||\,\vec{b}\,||^2+2\,\vec{x}\cdot\vec{y}.$ We thus obtain $ \vec{x}\cdot\vec{y} = \frac{1}{2}\left( \vec{a}\cdot\vec{b} + \overline{\vec{a}\cdot\vec{b}} \right). $
$\quad$ But observe that $\vec{a}\cdot\vec{b} = \alpha + i\beta$ for some $\alpha,\beta\in\mathbb{R}$. Then $ \vec{a}\cdot\vec{b} + \overline{\vec{a}\cdot\vec{b}} = (\alpha + i\beta)+(\alpha-i\beta) = 2\alpha = 2\text{Re}\,(\vec{a}\cdot\vec{b}). $ Hence, $ \vec{x}\cdot\vec{y} = \frac{1}{2}\left( \vec{a}\cdot\vec{b} + \overline{\vec{a}\cdot\vec{b}} \right) = \text{Re}\,(\vec{a}\cdot\vec{b}). $
And thus we finally have $ \cos\theta = \frac{\text{Re}\,(\vec{a}\cdot\vec{b})}{||\,\vec{a}\,||\,||\,\vec{b}\,||} \ . $ Therefore, $ \theta = \arccos \frac{\text{Re}\,(\vec{a}\cdot\vec{b})}{||\,\vec{a}\,||\,||\,\vec{b}\,||}. $