= Solution
Let $r=\operatorname{tr}C$, $s=\operatorname{tr}D$, and $\beta=(1-i)/2$. Their <traces> are real by part c, and $\Phi(C)=\beta rI+iC$. Thus $\operatorname{tr}\Phi(C)=(2\beta+i)r=r$. Expanding the product and using $2\beta^2+2i\beta=1$ gives
$$
\operatorname{tr}(\Phi(C)\Phi(D))=rs-\operatorname{tr}(CD).
$$
Therefore
$$
\boxed{b(\Phi(C),\Phi(D))
=\frac12[rs-\operatorname{tr}(CD)-rs]
=-\frac12\operatorname{tr}(CD)=c(C,D).}
$$
In 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 \b[<rank> $4$, inertia $(3,1)$], or numerical signature $2$, agreeing with $b$ under the real isomorphism.
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