Isometries of $\mathbb C^2$ I have studied isometries of the Euclidean plane $\mathbb R^2$ but am trying to understand what happens when we study $\mathbb C^2$ (over $\mathbb C$) instead.
In particular, I would like to see if a certain way of decomposing (uniquely) an isometry of the plane has some nice analogue in $\mathbb C^2$. 
To this end, I decided to generalize the notions of translation, rotation (about $\vec 0$), and reflection about the $x$-axis to get the following conjecture.

Given an isometry $\mathcal I$ of $\mathbb C^2$ (under the metric induced by the norm $\|(z,w)\|^2=z\bar{z}+w\bar{w}$), we have that
  $\mathcal I= T_{\vec v}\circ O\circ R$ for some translation $T$ by $\vec v$, some
  orthogonal transformation $O$, and some map $R$ which denotes either
  the reflection $F:(z,w)\mapsto (z, \bar{w})$ or the identity $I$.

My initial thought-process is as follows. Set $\vec x =\mathcal I(\vec 0)$. Then the translation in the desired decomposition would be $T_{\vec x}$. Moreover, notice that the map $A:=T_{-\vec x}\circ \mathcal I$ is linear. If $\det A>0$, then set $O=A$. Otherwise, we need $O$ to be something else. An intuitive choice is to set $O=AF$ in this case. But $F$ is not linear considered as map on $\mathbb C^2$ over $\mathbb C$.
However, $F$ is linear as a map on the four-dimensional space $\mathbb R^4$--that is, as the map $(a,b,c,d)\mapsto (a,b,-c,-d)$. And this should be fine, as I think that we can adjust the other maps easily to maps on $\mathbb R^4$.
Now, here are my questions. 
(1) Since I want $O$ to be orthogonal, is there a relatively easy way to show that $A$ or $AF$ (depending on the case) is orthogonal? I am unsure how to write down the matrix representations of these maps.
(2) And of course, I haven't shown the details of the actual proof of my conjecture. I just have been able to outline what I think should be the case. But I see no way of getting from the fact that $\mathcal I$ is distance-preserving to the fact that it can be decomposed in the way described. I have vaguely tried to play around with the eigenvalues for $O$, which I know should be complex conjugate pairs with norm $1$. But is it problematic if none of these is real, considering that we're technically using $\mathbb R^4$ and no longer $\mathbb C^2$? And to be honest, I am not sure anyway how studying the eigenvalues would help me with the proof ultimately. 
Could someone please help me with my efforts so far?
 A: $\newcommand{\Reals}{\mathbf{R}}\newcommand{\Cpx}{\mathbf{C}}$When you speak without qualification of an isometry with respect to the standard Hermitian norm, you're speaking of an isometry of Euclidean $4$-space. The story of Euclidean isometries is more-or-less uniform independently of dimension, so let's speak of Euclidean $n$-space with $n$ an arbitrary positive integer. Here are some useful factlets:


*

*If an isometry $F$ fixes the origin and the standard basis vectors, then $F$ is the identity.
Sketch of proof: This is clear if $n = 1$. To argue inductively on the dimension, decompose a point of $\Reals^{m+1}$ as $(x, x_{m}) = ((x_{1}, \dots, x_{m}), x_{m+1})$, and show $F$ preserves both $\|x\|$ (so $F(x) = x$ by the inductive hypothesis) and the $(m + 1)$th coordinate $x_{m+1}$.

*If an isometry $F$ fixes the origin, there exists a unique $n \times n$ (real) orthogonal matrix $A$ such that $F(x) = Ax$ for all $x$ in $\Reals^{n}$.
Sketch: By the isometry condition and the polarization identity $u \cdot v = \frac{1}{4}(\|u + v\|^{2} - \|u - v\|^{2})$, the points $F(e_{i})$ constitute an orthonormal set in $\Reals^{n}$. Let $A$ be the unique orthogonal matrix satisfying $F(e_{i}) = Ae_{i}$ for each $i$, and apply the first item to $A^{-1} \circ F$. (This answers your first question affirmatively.)

*If $F$ is an isometry, there exists a unique $n \times n$ orthogonal matrix and a unique $b$ in $\Reals^{n}$ such that
$$
F(x) = Ax + b\quad\text{for all $x$ in $\Reals^{n}$.}
$$
(You already have this proof. :)

*If $p$ is a point of $\Reals^{n}$ and $u$ is a unit vector, then reflection in the hyperplane through $p$ orthogonal to $u$ is the mapping
$$
R(x) = x - 2[(x - p) \cdot u] u,
$$
namely, $x$ minus "twice the $u$-component of $x - p$". For example, if $u = e_{n}$ and $p$ is the origin, then
$$
R(x_{1}, \dots, x_{n-1}, x_{n}) = (x_{1}, \dots, x_{n-1}, x_{n}) - 2x_{n} e_{n}
= (x_{1}, \dots, x_{n-1}, -x_{n}).
$$
The Euclidean group is generated by the set of reflections. In fact, if $F$ is a Euclidean isometry, there exist at most $(n + 1)$ reflections (not uniquely-defined!) whose composition is $F$.
A non-trivial translation is a composition of two reflections. (Why?)
A rotation is the composition of at most $n$ reflections, and is always the composition of an even number of reflections (because a reflection reverses orientation while a rotation preserves orientation). Thus, for example, every rotation of Euclidean $3$-space is a composition of at most two reflections, while a rotation of Euclidean $4$- or $5$-space is a composition of at most four reflections. (To prove this inductively, note that if $u$ and $v$ are arbitrary unit vectors, there exists a reflection exchanging $u$ and $v$.)

If, on the other hand, you want to speak only of Euclidean isometries of $\Cpx^{m} \simeq \Reals^{2m}$ for which the "rotation part" is complex-linear, you have the much smaller "holomorphic isometry group" corresponding to the unitary group $U(m) \subset SO(2m)$.
Thinking of $\Cpx^{2}$, the set of complex lines (a special class of real $2$-plane) is the well-known complex projective line, topologically a $2$-sphere, while the set of (oriented) real $2$-planes is topologically a product $S^{2} \times S^{2}$. This shows geometrically how small $U(2)$ is compared to $SO(4)$, 
