I was doing a physics class and the following analysis was done. It is about first order approximation of a function. The function is given by:$$f(x,y) = 1 + 2e^{i\frac{3}{2}x}\cos\big{(}\frac{\sqrt{3}}{2}y\big{)} $$This function has zeros at points $x = \frac{2\pi}{3}$ and $y = \pm \frac{2\pi}{3\sqrt{3}}$. From now on, let me write $\vec{x} = (x,y)$ and $\vec{a}_{\pm} = \big{(}\frac{2\pi}{3}, \pm \frac{2\pi}{3\sqrt{3}}\big{)}$.
The first step is to use a first order Taylor expansion. The professor wrote:$$f(\vec{x} + \vec{a}_{\pm}) = \vec{x}\cdot (\nabla f)(\vec{a}_{\pm}) \tag{1} $$since $f(\vec{a}_{\pm}) = 0$. Then, we can calculate the gradient:$$(\nabla f)(\vec{a}_{\pm}) = \big{(}-\frac{3}{2}i, \pm \frac{3}{2}\big{)}$$and we obtain:$$f(\vec{x}+\vec{a}_{\pm}) = -\frac{3}{2}ix \pm \frac{3}{2}y \Rightarrow |f(\vec{x}+\vec{a}_{\pm})| = \frac{3}{2}|\vec{x}|$$Finally, the main conclusion is:$$\frac{d}{d|\vec{x}|}f(\vec{x}+\vec{a}_{\pm}) = \frac{3}{2} \tag{2}$$
I am trying to justify these steps rigorously, as I think this is possible. But I have trouble justifying relations (1) and (2). First, to justify relation (1) I believe I need to use Taylor's expansion with remainder, so I would have to add to (1) an error term of order $\vec{x}^{2}$. However, since this is a complex function of real variable, I don't know exactly how to do it. I am confused with the formulas I know, because I only heard of Taylor series for real-valued functions of real variable or complex-valued functions of complex variable.
The second step, probably most difficult to justify is (2). If I introduce an error term in (1), it will probably carry out and an error term will also appear in (2). Then I would have to take the derivative with respect to $|\vec{x}|$. Since I did not accomplish the task of justifying (1), I still can't justify (2) too.
Can someone help me?