Intuitive explanation why complex differentiation is not linear over real and imaginary part Suppose we have a complex valued function $f = u + iv \colon U \to \mathbb C$ defined on an open subset $U$ of $\mathbb C$, which is holomorphic.
I was asked why it is wrong to use the linearity of the differentiation operator to write
$$
\frac{d}{dz} f(z) = \frac{d}{dz} (u(z) + iv(z)) = \frac{d}{dz} u(z) + i \frac{d}{dz} v(z).
$$
I explained this by proving that a real - valued function that is holomorphic must be constant, and so for $\frac{d}{dz} u$ and $\frac{d}{dz} v$ to exist $u$ and $v$ must be constant in $z$. However $f$ is not necessarily constant so applying the linearity in this way is invalid. I also showed that the real - and imaginary part functions $\Re()$ and $\Im()$ are not holomorphic.
Intuitively I think that real - valued and complex - valued functions are a different 'thing' and the $\frac{d}{dz}$ operator lives in the 'complex analysis world', so even though it is linear over complex functions it does not make sense to use it in the context of real - valued functions. However I was wondering whether there is a clearer or perhaps deeper explanation to this.

I would say that the formula is correct, but you need to specify that $u=u(z, \overline{z})$ and $v=v(z, \overline{z})$ and treat $z$ and $\overline{z}$ as two independent coordinates. You are indeed using complex coordinates to parametrize a real space: you are on $\mathbb{C}$ but you gave away the complex structure so you really are on $\mathbb{R}^2$. You need two independent coordinates to parametrize it. 
For example, $f(z)=z$ gives $u=\frac{1}{2}(z+\overline{z})$ and $v=\frac{1}{2i}(z-\overline{z})$. The formula is now correct: 
\begin{equation}
\begin{split}
1=\frac{dz}{dz} = \frac{\partial}{\partial z} \left( \frac{1}{2}(z+\overline{z})\right) + \frac{\partial}{\partial z} i\left(\frac{1}{2i}(z-\overline{z})\right) =1.
\end{split}
\end{equation}
See also Wirtinger derivatives. 
