why does $\frac{df}{dz}=\frac{1}{2}\left ( \frac{\partial f}{\partial x}-i\frac{\partial f}{\partial y}\right )$, why the $\frac{1}{2}$? I have trouble seeing where the $\frac{1}{2}$ comes from in $$\frac{df}{dz}=\frac{1}{2}\left ( \frac{\partial f}{\partial x}-i\frac{\partial f}{\partial y}\right )$$
For a change of variables $z=x+iy$ we have $$\frac{df}{dz}= \ \frac{\partial f}{\partial x}\frac{\partial x}{\partial z}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial z}$$ and $\frac{\partial x}{\partial z}=1$ and $\frac{\partial y}{\partial z}=-i$. Therefore we have the above but without the $\frac{1}{2}$. I've seen someone derive the correct expression by including the change of variables for $\overline{z}$ however I don't see how that is necessary, it should work without, right? I don't know what I am missing?
 A: Well it starts with the general formula for differential $$df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy.$$ In complex analysis we prefer to use $dz$ and $d\overline{z}$. So using, $dx = \dfrac{dz+d\overline{z}}{2}$ and $dy = \dfrac{dz-d\overline{z}}{2i},$ one finally gets $$df = \frac{1}{2}\left ( \frac{\partial f}{\partial x}-i\frac{\partial f}{\partial y}\right ) dz + \frac{1}{2}\left ( \frac{\partial f}{\partial x}+i\frac{\partial f}{\partial y}\right )d\overline{z}.$$ Now it is natural to define $\frac{\partial}{\partial z}$ to be $\frac{1}{2}\left ( \frac{\partial }{\partial x}-i\frac{\partial }{\partial y}\right )$ and $\frac{\partial}{\partial \overline{z}}$ to be $\frac{1}{2}\left ( \frac{\partial }{\partial x}+i\frac{\partial }{\partial y}\right )$. Thanks to that we have the beautiful formula $$df = \frac{\partial f}{\partial z} dz + \frac{\partial f}{\partial \overline{z}} d\overline{z}.$$
A: It is just a convention, there isn't a deep mathematical reason in the factor $\frac{1}{2}$. 
I think the convention comes from $dz=dx+idy$(Usually we prefer $n$-forms rather than $n$-tensor of tangent vectors, aren't we?) and then taking dual basis. 
