 Solution

Let $r=\mathrm{tr}C$, $s=\mathrm{tr}D$, and $\beta =(1-i)/2$. Their traces are real by part c, and $\Phi (C)=\beta rI+iC$. Thus $\mathrm{tr}\Phi (C)=(2\beta +i)r=r$. Expanding the product and using $2{\beta }^2+2i\beta =1$ givesThereforeIn particular the trace expression is real and symmetric on $Q$. Under the coordinates $C=aI+i\mathbf{v}\cdot \mathbf{A}$, $D=dI+i\mathbf{w}\cdot \mathbf{A}$, it is $-ad+\mathbf{v}\cdot \mathbf{w}$. Thus the trace form on scalar-plus-skew-Hermitian two-by-two matrices has rank $4$, inertia $(3,1)$, or numerical signature $2$, agreeing with $b$ under the real isomorphism.