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Without the usual math (i.e. the usual algebra you use to define these things), why should we want to define

$$\frac{d}{dz} = \frac{1}{2}(\frac{d}{dx}-i\frac{d}{dy})$$ $$\frac{d}{d\bar{z}} = \frac{1}{2}(\frac{d}{dx}+i\frac{d}{dy})$$

I keep forgetting the $\tfrac{1}{2}$ and keep forgetting the minus signs.

If we re-write $-i$ as $-i=\tfrac{1}{i}$ we can rewrite the above as

$$\frac{d}{dz} = \frac{1}{2}(\frac{d}{dx}+\frac{1}{i}\frac{d}{dy})$$ $$\frac{d}{d\bar{z}} = \frac{1}{2}(\frac{d}{dx}-\frac{1}{i}\frac{d}{dy})$$

which is a bit more intuitive since $\bar{z}$ is on the L.H.S. & there's a corresponding minus R.H.S. etc... but this still isn't intuitive enough - we can incorrectly derive them using something stupid like

$$\frac{d}{dz} = \frac{d}{d(x+iy)} = \frac{d}{dx} + \frac{d}{d(iy)} = \frac{d}{dx} + \frac{1}{i} \frac{d}{dy}$$

Is there a way to make this a bit nicer/logical and to include to $2$?

$$\frac{d}{dz} = \frac{d}{d[2(x+iy)]} = ...$$

Maybe with pictures?

Community
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bolbteppa
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    $\lim\limits_{w\to 0} \frac{f(z+w)-f(z)}{w}$ is (if it exists) the average of $\lim_{h\to 0} \frac{f(z+h)-f(z)}{h}$ and $\frac{1}{i} \lim_{k\to 0} \frac{f(z+ik)-f(z)}{k}$, where $w$ is complex, and $h,k$ are real. – Daniel Fischer Oct 12 '14 at 18:33
  • Related: http://math.stackexchange.com/questions/314863/what-is-the-intuition-behind-the-wirtinger-derivatives and http://mathoverflow.net/questions/92624/why-partial-and-bar-partial-defined-in-that-way-the-wirtinger-derivativ – echinodermata Oct 12 '14 at 20:32

2 Answers2

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The derivative $\displaystyle\frac{\partial}{\partial z}$ is completely fixed by linearity and two further requirements:

  • it should kill $\bar z=x-iy$ (this tells that $\partial_z\sim \partial_x-i\partial_y$)

  • $\partial_zz=1$ (this fixes the $\frac12$-factor).

Start wearing purple
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4

When you look at derivations, you look at the dual space of differential, you are not making some sort of funny division between $d$ and $dz$. If you consider the map sending $\{dx,dy\}$ to $\{dz,d\bar{z}\}$, written with respect to the basis $\{dx,dy\}$, you get $$A=\begin{pmatrix}1&1\\i&-i\end{pmatrix}$$ Now you want a map $B$ on the dual such that $\langle v, w^*\rangle=\langle Av, B^*w^*\rangle$ and so $B^{-1}=A^t$. Therefore $$B=\frac{1}{2}\begin{pmatrix}1& 1\\-i&i\end{pmatrix}\;.$$

Long story short, you are looking at a basis change on the dual space, so the related matrix is the transpose of the inverse of the original basis change.

wisefool
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