Consider $c,b \in \mathbb{C}$ and $f: \mathbb{C} \mapsto \mathbb{R}$, $$f(c) = bc' + b'c + cc'$$ is there only one extrema of $f$ corresponds to $$c^* = -b$$? where, $'$ means complex conjugate.
What I thought
since $df(c)/dc = b + b' + c + c' = 0$, therefore any $c \in \mathbb{C}$ such that $Re(b) + Re(c) = 0 $ will suffice. But I am very worried about what I thought which is so weird.
Writing $c=x + y i$, the hessian of your function is the identity matrix. Since that matrix is positive definite, the function is strictly convex in $x$ and $y$. It therefore only has one optimum.
Additionally writing $b=d+ei$, you get $f(c) = 2dx+2ey+x^2+y^2$. The derivative with respect to $x$ is $2d+2x$, so $x=d$ (this part you had). The derivative with respect to $y$ is $2e+2y$, so $y=e$ (this you did not have yet).