Matrices made of gluing $\begin{pmatrix} z & iw \\ i \bar w & \bar z \end{pmatrix}$ blocks have a real determinant 
Prove that matrices made entirely of blocks of the form $\begin{pmatrix}
z & iw \\ 
i \bar w & \bar z
\end{pmatrix}$ have a real determinant.

For example, we claim $$\Delta=\det \begin{pmatrix}
z_1 & iw_1 &z_2  & i \bar w_1\\ 
i \bar w_1 &\bar z_1  & i\bar w_2 & \bar z_2\\ 
z_3 & iw_3 &z_4  & i \bar w_4\\ 
i \bar w_3 &\bar z_3  & i\bar w_4 & \bar z_4
\end{pmatrix} \in \mathbb{R}\tag{*}$$
for all $z_i,w_i \in \mathbb{C}$.
Why this is interesting
If we start with an arbitrary square matrix $A$ over the quaternions $\mathbb{H}$, and send each element of the form $a+bi+cj+dk$ to the matrix $\begin{pmatrix}
a+bi & c+di \\ 
-c+di & a-bi
\end{pmatrix}$, we can define the determinant of $A$ as the determinant of the matrix obtained after expanding $A$ to a complex matrix in that way. 
Turns out this is the "right" choice, in conserving important rules like invertibility being equivalent to nonzero determinant. So in other words, we need to: 

Prove all quaternionic matrices have a real determinant.

For a single block the claim is obvious, but it is not clear how to reduce the general case to it.
 A: To see that $\Delta\in\mathbb{R}$ let us take the complex conjugate of it and show that $\overline{\Delta}=\Delta$ in (*). It will be clear what happens with all the blocks by looking at the first block only.
$$
\overline{\Delta}=\det\overline{\begin{pmatrix}
z & iw \\ 
i \bar w & \bar z
\end{pmatrix}}=
\det\begin{pmatrix}
\bar z & -i\bar w \\ 
-iw & z
\end{pmatrix}.
$$
Now we switch every odd row (in total $n$ many) with the next even row and then every odd column (also $n$ many) with the next even one. The determinant changes the sign $2n$ times, i.e. keeps the same one.
$$
\det\begin{pmatrix}
\bar z & -i\bar w \\ 
-iw & z
\end{pmatrix}=-\det\begin{pmatrix}
-iw & z\\
\bar z & -i\bar w  
\end{pmatrix}=\det\begin{pmatrix}
z & -iw \\ 
-i\bar w & \bar z
\end{pmatrix}.
$$
Now multiply every odd column and then every odd row by $-1$. The determinant is multiplied by $(-1)^{2n}=1$
$$
\det\begin{pmatrix}
z & -iw \\ 
-i\bar w & \bar z
\end{pmatrix}=-\det\begin{pmatrix}
-z & -iw \\ 
i\bar w & \bar z
\end{pmatrix}=\det\begin{pmatrix}
z & iw \\ 
i\bar w & \bar z
\end{pmatrix}=\Delta.
$$
Since $\overline{\Delta}=\Delta$ it is real.
