Let $z=a+bi \in \mathbb C$ be a complex number with $a,b \in \mathbb R$, and define $$ M(z)= \begin{pmatrix} a & -b\\ b & a \end{pmatrix}. $$ Show that det($M(z)$) = $|z|, \forall z \in \mathbb C$.
I found the determinant to be $a^2+b^2$, but I'm not sure how to incorporate the $z=a+bi \in \mathbb C$ part into the proof.
As stated, the problem is wrong. If you consider $a=b=1$, then $\det M=2$, while $|1+i|=\sqrt2. $ The right equality would be $$\det M=|z|^2. $$