Gradient of complex product

Question

So suppose I have a a complex affine (multivariable) function defined as follows: $$f(x) = a_0 + a_1x_1 + a_2 x_2 + \ldots + a_n x_n$$ where all the $a$'s are complex numbers, all the $x$'s are real numbers. I am interested in finding the first order Taylor series of the function: $$g(x) = f^{\star}(x)f(x)$$ around an operating point $c = [c_1, c_2, \dots,c_n]^T$ (where the star signifies the complex conjugate). Is there a way to do this in terms of gradients? For example, would it be correct to assert the following

$$f^{\star}(x)f(x) \approx f^{\star}(c)f(c) + D_g(c)(x-c)$$ where $$D_g(c) = \left[\frac{dg(x)}{dx_1}(c), \frac{dg(x)}{dx_2}(c),\ldots, \frac{dg(x)}{dx_n}(c)\right]$$

Also, for complex functions, does the product rule still hold? For example: $$\frac{dg(x)}{dx_1} = f^{\star}(x)\frac{df(x)}{dx_1} + f(x)\frac{df^{\star}(x)}{dx_1}$$

Your help is much appreciated!!

Answer 1

Generally speaking, almost all usual operations are allowed. Let us be more precise.

In order to have a clear view of what you can do or not do with a function $f:\mathbb{R^n}\rightarrow \mathbb{C}$, convert it into an equivalent function $(f_r,f_i):\mathbb{R^n}\rightarrow \mathbb{R^2}$ obtained by taking $f_r=Re(f)$, $f_i=Im(f)$, and convert as well the operations you want to do.

For example, is $f^{\star}:x \rightarrow f^{\star}(x)=f_r(x)-if_i(x)$ a differentiable function ? We convert it into the issue of differentiability of function $x \rightarrow (f_r(x),-f_i(x))$ from $\mathbb{R^n}$ to $\mathbb{R^2}$. And the answer is yes.

In particular, Taylor expansions are allowed as long as they make sense for corresponding functions $\mathbb{R^n} \rightarrow \mathbb{R^2}.$

Answer 2

You have $f:\mathbb{R}^n\rightarrow\mathbb{C}$, so write $f=u+iv$, where $u,v:\mathbb{R}^n\rightarrow\mathbb{R}$. Then $g=(u-iv)(u+iv)=u^2+v^2$, so $g:\mathbb{R}^n\rightarrow\mathbb{R}$. Then, by Taylor's theorem in several variables, $$g(x)\approx g(c)+D_g(c)(x-c)$$ as you asserted. In fact, as $g$ is a real polynomial of degree 2 in $n$ variables, $$ g(x)=g(c)+D_g(c)(x-c)+(x-c)^TH_g(c)(x-c),$$ where $H_g$ is the Hessian matrix of $g$.

For the chain rule (I will denote with a ' the derivative with respect to $x_1$): $$ g'=2(uu'+vv'), $$ $$ f^*f'+(f^*)'f=(u-iv)(u'+iv')+(u+iv)(u'-iv')=2(uu'+vv'), $$ so in this case the chain rule holds.

REMARK: For a general function $g:\mathbb{R}^n\rightarrow\mathbb{C}$, Taylor's theorem does not hold. For instance, if $n=1$ and $g(x)=e^{ix}$, then $g(2\pi)-g(0)=0$, but $|g'|=1$.

REMARK: For general functions $g,h:\mathbb{R}\rightarrow\mathbb{C}$, the chain rule $(hg)'=h'g+g'h$ remains true.