Prove a unique real matrix exists to denote complex numbers in the set of Cauchy–Riemann matrices One model for the complex numbers $\mathbb{C}$ uses the set of Cauchy–Riemann matrices 
$CR:=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ such that $a,b,c,d \in \mathbb{R}, a=d, b+c=0$ with matrix addition and matrix multiplication corresponding to $+$ and $×$ in $\mathbb{C}$. We say that a matrix$ 
\begin{pmatrix}a&b\\c&d\end{pmatrix} \in$ CR is real if $b=c=0$ and imaginary if $a=d=0$, we write $i:=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$
$(a)$ Show that for every imaginary $z ∈$ CR there is a unique real matrix $u ∈$ CR such that $z = u · i.$
$(b)$ Show that for every $z ∈$ CR there are unique real matrices $u, v ∈$ CR such that $z = u + v · i$.
I believe for both parts of the question, I am missing a key aspect of proving uniqueness. Any help would be appreciated 
 A: Some intuition on CR matrices may be gained via inspection of the following simple matrix equations (1)-(3), easily verified:
$\begin{bmatrix} a & 0 \\ 0 & a \end{bmatrix} \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & -a \\ a & 0 \end{bmatrix}; \tag 1$
$\begin{bmatrix} 0 & -a \\ a & 0 \end{bmatrix}\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} -a & 0 \\ 0 & -a \end{bmatrix}; \tag 2$
$\begin{bmatrix} 0 & -a \\ a & 0 \end{bmatrix}\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} = \begin{bmatrix} a & 0 \\ 0 & a \end{bmatrix}; \tag 3$
we have
$i = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}; \tag 4$
$i^2 = -I; \tag 5$
$(-i)i = -i^2 = I; \tag 6$
$i^{-1} = -i. \tag 7$
Note that (1) shows that for every imaginary $z \in CR$ there is a real $u \in CR$ such that
$ui = z; \tag 8$
if 
$yi = z, \tag 9$
then in light of (5),
$-y = -yI = yi^2 = zi, \tag{10}$
or
$y = -zi, \tag{11}$
which shows that there is precisely one $u$ satisfying (8); thus is item (a) resolved.
We next observe that the only $w \in CR$ which is both real and imaginary is $0$, for if $w$ is real then
$w = \begin{bmatrix} a & 0 \\ 0 & a \end{bmatrix} \tag{12}$
for some $a \in \Bbb R$, and since $w$ is also imaginary, we have
$w = \begin{bmatrix} 0 & b \\ -b & 0 \end{bmatrix} \tag{13}$
for some $b \in \Bbb R$; equating these two forms of $w$ yields
$\begin{bmatrix} a & 0 \\ 0 & a \end{bmatrix} = \begin{bmatrix} 0 & b \\ -b & 0 \end{bmatrix}; \tag{14}$
comparing entries of these two matrices indicates that
$a = b = 0, \tag{15}$
whence
$w = 0, \tag{16}$
as asserted.
Now any $z \in CR$ may be written
$z = \begin{bmatrix} a & -b \\ b & a \end{bmatrix} \tag{17}$
for some $a, b \in \Bbb R$; we have
$z = \begin{bmatrix} a & 0 \\ 0 & a \end{bmatrix} + \begin{bmatrix} 0 & -b \\ b & 0 \end{bmatrix} = \begin{bmatrix} a & 0 \\ 0 & a \end{bmatrix} + \begin{bmatrix} b & 0 \\ 0 & b \end{bmatrix}\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}; \tag{18}$
we denote the real matrices occurring in this equation by $u$ and $v$:
$u = \begin{bmatrix} a & 0 \\ 0 & a \end{bmatrix}, \tag{19}$
$v = \begin{bmatrix} b & 0 \\ 0 & b \end{bmatrix}, \tag{20}$
then using (4) we have
$z = u + vi; \tag{21}$
if there were two other real matrices $u'$ and $v'$ such that
$z = u' + v'i, \tag{22}$
then
$u' + v'i = u + vi, \tag{23}$
or
$u' - u = vi - v'i = (v - v')i; \tag{24}$
since $u' - u$ and $v - v'$ are both real, $(v - v')i$ is imaginary in light of (1), and thus by what we have just seen in (12)-(18) it follows that
$u' - u = (v - v')i = 0, \tag{25}$
whence
$u' = u, \tag{26}$
and again by virtue of (5),
$v - v' = (v - v')I = -(v - v')i^2 = (-(v - v')i)i = 0, \tag{27}$
and thus
$v = v' \tag{28}$
as well. Thus the uniqueness of $u$, $v$ as in (21) is established, and we have dispensed with item (b).
A: Since $i^2=-I_2$, For (a) the required $u$ is $$-zi=-\left(\begin{array}{cc}
0 & b_{z}\\
-b_{z} & 0
\end{array}\right)\left(\begin{array}{cc}
0 & -1\\
1 & 0
\end{array}\right)=\left(\begin{array}{cc}
-b_{z} & 0\\
0 & -b_{z}
\end{array}\right).$$For (b) note that$$u+vi=\left(\begin{array}{cc}
a_{u} & 0\\
0 & a_{u}
\end{array}\right)+\left(\begin{array}{cc}
a_{v} & 0\\
0 & a_{v}
\end{array}\right)\left(\begin{array}{cc}
0 & -1\\
1 & 0
\end{array}\right)=\left(\begin{array}{cc}
a_{u} & -a_{v}\\
a_{v} & a_{u}
\end{array}\right),$$which agrees with $\left(\begin{array}{cc}
a_{z} & b_{z}\\
-b_{z} & a_{z}
\end{array}\right)$ iff $a_{u}=a_{z},\,a_{v}=-b_{z}$.
