An equation in a commutative field ` Consider :
       $$I=\left( \begin{matrix} 1& 0\\ 0& 1\end{matrix} \right);J=\left( \begin{matrix} 1& -1\\ 3& -1\end{matrix} \right)$$
  $$From:   M_{2}\left( \mathbb{R} \right)$$
And the set : $$E=\left\{M\left( a,b\right)=aI+b\dfrac {\sqrt {2}} {2}J;\left( a;b\right) \in \mathbb{R} ^{2} \right\}$$
Solve in E the following equation : $$X^{4}=I$$
Additional details : 
         - $$f\left( a+bi\right)=M\left( a;b\right)$$  "f" is an isomorphism  $$ \left( \mathbb{C} ,⨯\right) \rightarrow \left( E,⨯\right)  $$ 
          - E is a commutative field .
 A: Since $f$ is an isomorphism, we have
$$
I=M(1,0)=f(1),\quad J=\sqrt{2}M(0,1)=\sqrt{2}f(i);
$$
in particular
$$
J^2=2f(i)f(i)=2f(-1)=-2I
$$
Solving the equation
$$\tag{1}
X^4=I
$$
is therefore equivalent to solving the equation
$$\tag{2}
x^4=1
$$
with $x=f^{-1}(X)$.
Since the solutions of $(2)$ are
$$
\pm1,\pm i,
$$
the solutions of $(1)$ are therefore
$$
f(\pm1)=\pm f(1)=\pm I,\quad f(\pm i)=\pm f(i)=\pm\dfrac{1}{\sqrt{2}}J
$$
A: I'm not sure that "f is an isomorphism" was an input of the problem as Mercy King assumed, but rather a hint to be proved, as well as $E$ being a field.
$J^2=\begin{pmatrix} 1 & -1 \\ 3 & -1 \end{pmatrix}\begin{pmatrix} 1 & -1 \\ 3 & -1 \end{pmatrix}=\begin{pmatrix} -2 & 0 \\ 0 & -2 \end{pmatrix}=-2I$
$M(a,b)M(c,d)=(aI+\frac b{\sqrt{2}}J)(cI+\frac d{\sqrt{2}}J)=(ac-bd)I+(\frac{ad+bc}{\sqrt{2}})J=M(ac-bd,ad+bc)$
So if we define $\begin{cases} z_1=a+ib & f(z_1)=M(a,b)\\ z_2=c+id & f(z_2)=M(c,d)\end{cases}$
we get $f(z_1)f(z_2)=f((ac-bd)+i(ad+bc))=f(z_1z_2)$ and $f$ is an isomorphism in $(\mathbb C,\times)$.
Since $M(a,b)$ definition is linear in $a,b$ then $f$ is also an isomorphism in $(\mathbb C,+)$.
Neutral element for $(\mathbb C,+)$ is $0$ which correspond to $M(0,0)=0$ also neutral element for matrices addition.
Neutral element for $(\mathbb C,\times)$ is $1$ which correspond to $M(1,0)=I$ also neutral element for matrices multiplication.
So we can transport the field structure between $(\mathbb C,+,\times)$ and $(E,+,\times)$ with this $f$.
For the solving of $X^4=I$ see Mercy King's answer.
