6
$\begingroup$

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?

$\endgroup$
4
$\begingroup$

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 $$df = \frac{\partial f}{\partial z} dz + \frac{\partial f}{\partial \overline{z}} d\overline{z}.$$

$\endgroup$
  • $\begingroup$ How can I interpret $\frac{\partial f}{\partial x} dx$? What does $dx$ mean in that expression? $\endgroup$ – Hendrra Mar 27 '18 at 16:58
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.