Differentiation by a complex vector

Complex numbers frequently arise in physical problems, particularly with Fourier series. Let us extend the multivariable least-squares theory to the use of complex-valued unknowns $\bold x$.First recall how complex numbers were handled with single-variable least squares, i.e., as in the discussion leading up to equation (). Use prime, such as $\bold x'$, to denote the complex conjugate of the transposed vector $\bold x$.Now write the positive quadratic form as  

$$Q(\bold x', \bold x) \eq (\bold B\bold x - \bold d)' (\bold ... ...) \eq (\bold x' \bold B' - \bold d') (\bold B\bold x - \bold d)$$ (25)

In chapter (after equation ()), we minimized a quadratic form $Q(\bar X,X)$by setting to zero both $\partial Q/\partial \bar X$ and $\partial Q/\partial X$.We noted that only one of $\partial Q/\partial \bar X$ and $\partial Q/\partial X$is necessary because they are conjugates of each other. Now take the derivative of Q with respect to the (possibly complex, row) vector $\bold x'$.Notice that $\partial Q/\partial \bold x'$ is the complex conjugate transpose of $\partial Q/\partial \bold x$.Thus, setting one to zero sets the other also to zero. Setting $\partial Q/\partial \bold x' =\bold 0$ gives the normal equations:   

$$\bold 0 \eq {\partial Q \over \partial \bold x'} \eq \bold B' (\bold B\bold x - \bold d)$$ (26)

The result is merely the complex form of our earlier result (14). Therefore, differentiating by a complex vector is an abstract concept, but it gives the same set of equations as differentiating by each scalar component, and it saves much clutter.