Angle between two vectors?

I have been taught that the angle between two vectors is supposed to be their inner product. However, the book I’m reading states:

Recall that the angle between two vectors $u = (u_0,\ldots,u_{n−1})$ and $v = (v_0,\ldots, v_{n−1})$ in $\mathbb{C}^n$ (the complex plane) is just a scaling factor times their inner product.

What is a “scaling factor”?

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Your statement that

the angle between two vectors is supposed to be their inner product

is incorrect, as is the statement from the book. On the Wikipedia page on the dot product, you can see the correct formula for the angle between two complex vectors $u$ and $v$ (thanks to Henry for catching the earlier mistake):
$$\theta=\arccos\left(\frac{\operatorname{Re}(u\cdot v)}{\|u\|\|v\|}\right)$$
where the inner product $u\cdot v$ is defined to be
$$u\cdot v=\sum_{k=0}^{n-1} u_k\overline{v_k}$$
I would guess that perhaps the intended meaning of the “scaling factor” is as follows: when $u$ and $v$ are unit vectors, we have
$$\cos(\theta)=\operatorname{Re}(u\cdot v)$$
while when $u$ and $v$ are arbitrary non-zero vectors, we have
$$\cos(\theta)=\frac{\operatorname{Re}(u\cdot v)}{\|u\|\|v\|}$$
(the quantities $\|u\|$ and $\|v\|$ are both equal to $1$ when $u$ and $v$ are unit vectors). This would make $$\frac{1}{\|u\|\|v\|}$$
the “scaling factor”, though it is scaling the formula for the cosine of the angle, not the angle itself.

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,
$$
\theta = \arccos \frac{\text{Re}\,(\vec{a}\cdot\vec{b})}{||\,\vec{a}\,||\,||\,\vec{b}\,||}.
$$

The dot product, inner product or scalar product is defined as:
$$\vec u \cdot \vec v = u_1\cdot v_1 + \ldots + u_n\cdot v_n = \|\vec u\|\cdot\|\vec v\|\cdot\cos(\angle\vec u\vec v)$$
In other words, if the two vectors are unit vectors, their dot product is the cosine of the angle between them. To get to that point, simply normalize the two vectors:
$$\frac{\vec u}{\|\vec u\|}\cdot\frac{\vec v}{\|\vec v\|}=\frac{\vec u \cdot \vec v}{\|\vec u\|\cdot\|\vec v\|}=\cos(\angle\vec u\vec v)$$
Take the arcus cosine of the quotient and you get the actual angle.