Show a rotation is given by $v=e^{-i\theta }ue^{i\theta }$ I am currently reading Alan Macdonald's book on Linear and Geometric Algebra. I can't seem to figure out a line:

Why is line 4 true? The 3 previous lines make sense.
Thanks for any insights you may provide!
 A: I'd suggest popping in some specific vectors to get a feeling for things.  For example, suppose that $ \mathbf{i} = \mathbf{e}_1 \mathbf{e}_2 $,
so that $ \mathbf{u}_\parallel = a \mathbf{e}_1 + b \mathbf{e}_2 $, and $ \mathbf{u}_\perp = c \mathbf{e}_3 + \cdots $.
Since
$$\begin{aligned}   \mathbf{e}_1 e^{\mathbf{i} \alpha}    &= \mathbf{e}_1 \left( { \cos\alpha + \mathbf{e}_1 \mathbf{e}_2 \sin\alpha } \right) \\    &= \cos\alpha \mathbf{e}_1 + \sin\alpha\, \mathbf{e}_1 \mathbf{e}_1 \mathbf{e}_2 \\    &= \cos\alpha \mathbf{e}_1 - \sin\alpha\, \mathbf{e}_1 \mathbf{e}_2 \mathbf{e}_1 \\    &= \left( { \cos\alpha - \mathbf{i} \sin\alpha } \right) \mathbf{e}_1 \\    &= e^{-\mathbf{i} \alpha} \mathbf{e}_1,\end{aligned}$$
and
$$\begin{aligned}   \mathbf{e}_2 e^{\mathbf{i} \alpha}    &= \mathbf{e}_2 \left( { \cos\alpha + \mathbf{e}_1 \mathbf{e}_2 \sin\alpha } \right) \\    &= \cos\alpha \mathbf{e}_2 + \sin\alpha\, \mathbf{e}_2 \mathbf{e}_1 \mathbf{e}_2 \\    &= \left( { \cos\alpha + \mathbf{e}_2 \mathbf{e}_1 \sin\alpha } \right) \mathbf{e}_2 \\    &= \left( { \cos\alpha - \mathbf{i} \sin\alpha } \right) \mathbf{e}_2 \\    &= e^{-\mathbf{i} \alpha} \mathbf{e}_2.\end{aligned}$$
You can use a superposition argument to conclude that
$$   \mathbf{x} e^{\mathbf{i} \alpha} = e^{-\mathbf{i} \alpha} \mathbf{x},$$
for any $ \mathbf{x} \in \text{span} \left\{ {\mathbf{e}_1, \mathbf{e}_2} \right\} $.  In particular $ \mathbf{u}_\parallel e^{\mathbf{i}\theta/2} = e^{-\mathbf{i}\theta/2} \mathbf{u}_\parallel $.
For any component of the vector that is perpendicular to the plane of rotation, such as $ \mathbf{e}_3 $ for example, we have
$$\begin{aligned}   \mathbf{e}_3 e^{\mathbf{i} \alpha}    &= \mathbf{e}_3 \left( { \cos\alpha + \mathbf{e}_1 \mathbf{e}_2 \sin\alpha } \right) \\    &= \cos\alpha \mathbf{e}_3 + \sin\alpha\, \mathbf{e}_3 \mathbf{e}_1 \mathbf{e}_2 \\    &= \left( { \cos\alpha + \mathbf{i} \sin\alpha } \right) \mathbf{e}_3 \\    &= e^{\mathbf{i} \alpha} \mathbf{e}_3,\end{aligned}$$
since the sign changes twice as you permute the factors of the trivector (i.e.: $ \mathbf{e}_3 \mathbf{e}_1 \mathbf{e}_2 = - \mathbf{e}_1 \mathbf{e}_3 \mathbf{e}_2 = + \mathbf{e}_1 \mathbf{e}_2 \mathbf{e}_3 $ ).  Again, a superposition argument allows you to conclude that $ \mathbf{u}_\perp e^{\mathbf{i}\theta/2} = e^{\mathbf{i}\theta/2} \mathbf{u}_\perp $.
In short, a complex exponential for a rotation in the plane $ \mathbf{i} $ commutes with components of the vector that lie off the plane, but conjugate-commutes with components of the vector that lie in the plane.
These two facts allow you to make sense of what you've labelled line 4.
