For $q=z+jw$ where $z,w\in\mathbb{C}$, I'm given a map $M:\mathbb{C}^2\rightarrow M_{2\times2}(\mathbb{C})$ given by $$M(q)=\begin{pmatrix} z & \overline{w} \\ -w & \overline{z} \end{pmatrix}$$
and I want to show that $M(q_1q_2)=M(q_1)M(q_2)$. I can't seem to avoid doing a long computation, which is difficult to carry out. Is there a faster way that I'm missing?
No long computation: $$ \begin{bmatrix} z_1 & \bar{w}_1 \\ -w_1 & \bar{z}_1 \end{bmatrix} \begin{bmatrix} z_2 & \bar{w}_2 \\ -w_2 & \bar{z}_2 \end{bmatrix} = \begin{bmatrix} z_1z_2-\bar{w}_1w_2 & z_1\bar{w}_2+\bar{w}_1\bar{z}_2\\ -(z_2w_1+\bar{z}_1w_2) & \bar{z}_1\bar{z}_2-w_1\bar{w}_2 \\ \end{bmatrix} $$
Now observe that, for real $\alpha$ and $\beta$, $$ j(\alpha+i\beta)=j\alpha+ji\beta=j\alpha-ij\beta=(\alpha-i\beta)j $$ so $jw=\bar{w}j$, for $w\in\mathbb{C}$. Then \begin{align} (z_1+jw_1)(z_2+jw_2) &=z_1z_2+jw_1z_2+z_1jw_2+jw_1jw_2\\ &=z_1z_2+jw_1z_2+j\bar{z}_1w_2+j^2\bar{w}_1w_2\\ &=(z_1z_2-\bar{w}_1w_2)+j(z_2w_1+\bar{z}_1w_2) \end{align}