Difference between interpretation of $f(x, y)$ and $f(x + iy)$ when calculating the partial derivatives of $f$'s component functions Let $\Omega\subset\mathbb{C}$ and $f:\Omega\to\mathbb{C}$ be a differentiable mapping $f = u + iv$. Define $F(x, y) = (\mathfrak{Re}(f(x + iy)), \mathfrak{Im}(f(x + iy))$. Then isn't it true that $\nabla \cdot F = \frac{\partial u(x + iy)}{\partial x}\cdot 1 + \frac{\partial v(x + iy)}{\partial y}\cdot i$ as per the chain rule? However, if we were to view $x + iy \equiv (x, y)$, then we wouldn't have the coefficient $i$ with $\frac{\partial v(x + iy)}{\partial y}$. Difference between the two isomorphically same interpretations doesn't seem correct, but I am not sure where the fault lies.
 A: Not sure what you are asking.
Let's stick to the complex representation using $f=u+iv$. The chain rule gives me the two equations
\begin{align}
\frac{\partial}{\partial x}f&=\frac{\partial}{\partial x}u+i\frac{\partial}{\partial \color{red}{x}}v\,,\\
\frac{\partial}{\partial y}f&=\frac{\partial}{\partial\color{red}{y}}u+i\frac{\partial}{\partial y}v\,.
\end{align}
The matrix
$$
\begin{pmatrix}\frac{\partial}{\partial x}u&\frac{\partial}{\partial y}u\\
\frac{\partial}{\partial x}v&\frac{\partial}{\partial y}v\\
\end{pmatrix}
$$
is the Jacobian of $F$ and
$$
\nabla\cdot F=\frac{\partial}{\partial x}u+\frac{\partial}{\partial y}v
$$
(without $i$) is the divergence of $F$.
A: Let $f^{cc}: \mathbb{C} \to \mathbb{C}$ be a real differentiable function.
I will use superscripts of $cc$ to indicate $\mathbb{C} \to \mathbb{C}$, $cr$ to indicate $\mathbb{C} \to \mathbb{R}$ or $\mathbb{C} \to \mathbb{R}^2$, and $rr$ to indicate $\mathbb{R}^2 \to \mathbb{R}^2$ or $\mathbb{R}^2 \to \mathbb{R}$.
Let $f^{cc}(x+iy) = u^{cr}(x+iy) + v^{cr}(x+iy)$ where $u^{cr}, v^{cr}: \mathbb{C} \to \mathbb{R}$.
Define $u^{rr}, v^{rr}: \mathbb{R}^2 \to \mathbb{R}$ by $u^{rr}(x,y) = u^{cr}(x+iy)$ and $v^{rr}(x,y) = v^{cr}(x+iy)$.
Finally define $f^{rr}: \mathbb{R}^2 \to \mathbb{R}^2$ by $f^{rr}(x,y) = (u^{rr}(x,y), v^{rr}(x,y))$.
It seems to me that you are getting confused because you are not sure if there is a difference between $\frac{\partial v^{cr}}{\partial y}$ and $\frac{\partial v^{rr}}{\partial y}$.  I think you are confused because you are using the same name $v$ for these two different (but closely related!) functions.
The way you are using notation is leading you to believe that you need to use the chain rule, and introduce another factor of $i$ when you evaluate $\frac{\partial v^{cr}}{\partial y}$.  This is an illusion.  First of all:  it wouldn't make any sense.  The codomain of $v^{cr}$ is the real numbers.  Multiplying by $i$ in the codomain just doesn't make sense.
In fact (for any sensible definition of $\frac{\partial}{\partial y}$ as an operator on these different function spaces) we should have $$\frac{\partial v^{cr}}{\partial y} = \frac{\partial v^{rr}}{\partial y}$$.
$
\begin{align*}
\frac{\partial v^{cr}}{\partial y} \big|_{(a,b)} &= \lim_{h \to 0} \frac{v^{cr}(a+(b+h)i) - v^{cr}(a+bi)}{h}\\
&= \lim_{h \to 0} \frac{v^{rr}(a, b+h) - v^{cr}(a,b)}{h}\\
&= \frac{\partial v^{rr}}{\partial y} \big|_{(a,b)}
\end{align*}$
For this reason, most authors would just write $\frac{\partial v}{\partial y}$ instead of writing either $\frac{\partial v^{cr}}{\partial y}$ or $\frac{\partial v^{rr}}{\partial y}$ since these numbers are equal.
It is often hard to keep your head on straight when you need to keep shifting perspective between functions from/to $\mathbb{C}$ and functions from/to $\mathbb{R}^2$.  I hope that explicitly distinguishing between these has helped to resolve your confusion.
I commend you for trying to think through this.  Most texts in complex analysis do a very poor job of helping students think through these issues.  There is a lot of abuse of notation, and I think that this kind of thing often leads to real confusion.  Keep battling with it until it makes sense!

Edit:  Here are some additional comments (not directly related to your question) which I hope can clear up some other common related confusions.  Again, most textbooks focus so exclusively on holomorphic functions that they actually create confusion about what happens in the non-holomorphic case.  They also do not clarify these distinctions between $2n$ real dimensional and $n$ complex dimensional perspectives.  This comes back to bite you if you ever decide to do several complex variables.  So it is well worth understanding when studying one complex variable, in my opinion.
$Df^{cc} \big|_{a+bi}$ is a real-linear map.  The matrix of this real-linear map when I use the basis ${1,i}$ for both the domain and codomain is
$$\left.
\begin{bmatrix}
\frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}
\end{bmatrix}\right|_{(a,b)}
$$
Notice that the space of real-linear maps $\mathbb{R}Lin(\mathbb{C}, \mathbb{C})$ is a $4$ dimensional real vector space, but it is a 2 dimensional complex vector space.  We can naturally scale any real linear map $L: \mathbb{C} \to \mathbb{C}$ by a complex scalar just by pointwise multiplication.  The resulting map is still real linear. So $\mathbb{R}Lin(\mathbb{C}, \mathbb{C})$ is a 2 dimensional complex vector space.  A nice basis of this space is $\textrm{dz}$ which is just the identity map and $\textrm{d}\bar{z}$ which is complex conjugation.  Expressing a real linear map in this basis decomposes it into a complex linear part and a complex anti-linear part.
Given any real linear map $L: \mathbb{C} \to \mathbb{C}$ we can try to express it as a complex linear combination of $\textrm{d}z$ and $\textrm{d}\bar{z}$.  So we want $L(z) = a \textrm{d}z + b \textrm{d}\bar{z}$ for two complex numbers $a,b$.  Let's see what happens when we evaluate on the basis vectors $1$ and $i$:
\begin{align*}
L(1) = a \textrm{d}z (1) + b \textrm{d}\bar{z}(1)\\
L(i) = a \textrm{d}z (i) + b \textrm{d}\bar{z}(i)\\
\end{align*}
so
\begin{align*}
L(1) = a  + b\\
L(i) = ai - bi\\
\end{align*}
so
\begin{align*}
iL(1) = ai  + bi\\
L(i) = ai - bi\\
\end{align*}
\begin{align*}
iL(1) + L(i) = 2ai\\
iL(1) - L(i) = 2bi\\
\end{align*}
So
\begin{align*}
a = \frac{1}{2i}(iL(1)+L(i))\\
b = \frac{1}{2i}(iL(1)-L(i))\\
\end{align*}
Define $\frac{\partial f}{\partial x} = \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x}$ and  $\frac{\partial f}{\partial y} = \frac{\partial u}{\partial y} + i \frac{\partial v}{\partial y}$
Decomposing $Df^{cc}$ into complex linear and anti linear parts we get
$
\begin{align*}
Df^{cc} &= \frac{1}{2i}(i (\frac{\partial f}{\partial x}) + \frac{\partial f}{\partial y}) \textrm{d} z + \frac{1}{2i}(i (\frac{\partial f}{\partial x}) - \frac{\partial f}{\partial y}) \textrm{d} \bar{z}\\
&=\frac{1}{2}(\frac{\partial f}{\partial x} - i\frac{\partial f}{\partial y}) \textrm{d}z + \frac{1}{2}( \frac{\partial f}{\partial x} + i \frac{\partial f}{\partial y}) \textrm{d} \bar{z}
\end{align*}
$
This further motivates the definitions
$$\frac{\partial f}{\partial z} = \frac{1}{2}(\frac{\partial f}{\partial x} - i\frac{\partial f}{\partial y})$$
$$\frac{\partial f}{\partial \bar{z}} = \frac{1}{2}(\frac{\partial f}{\partial x} + i\frac{\partial f}{\partial y})$$
In other words, $\frac{\partial f}{\partial z}$ is the coefficient of the complex linear part of the real derivative while $\frac{\partial f}{\partial z}$ is the coefficient of the complex antilinear part of the real derivative.
For a function to be holomorphic we want that its derivative is complex linear.  This means we want its complex anti-linear part to vanish.  So we want
$$
\begin{align*}
\frac{\partial f}{\partial \bar{z}} &= 0\\
(\frac{\partial f}{\partial x} + i\frac{\partial f}{\partial y}) &= 0\\
\frac{\partial f}{\partial x} &= -i \frac{\partial f}{\partial y})\\
\frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x} &= -i (\frac{\partial u}{\partial y} + i \frac{\partial v}{\partial y})\\
\end{align*}
$$
so, equating real and imaginary parts,
$$
\begin{cases}
\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}\\
\frac{\partial v}{\partial x} = - \frac{\partial u}{\partial y}
\end{cases}
$$
which are the Cauchy-Riemann equations.
This is in harmony with the fact that a real linear map $L: \mathbb{C} \to \mathbb{C}$ is only complex linear if its matrix with respect to the basis ${1,i}$ is of the form
$$
\begin{bmatrix}
a & -b \\ b & a
\end{bmatrix}
$$
In other words, the Cauchy-Riemann equations just say that the real-derviative is complex linear.
