# How is the operator $\frac{\partial}{\partial z}$ defined?

My book says that $$\frac{\partial}{\partial z}:=\frac12\left(\frac{\partial}{\partial x}-i \frac{\partial}{\partial y}\right),$$ but I don't know why $$\frac{\partial}{\partial z}$$ should be defined that way. Does anyone have an idea why we should have this definition?

I have just done complex analysis at the moment - $$\frac{\partial}{\partial z}$$ seems to be related to multivariable complex functions, which I am not familiar with.

• Consider what the derivative should do to $z = x + iy$. – Cameron Williams Sep 27 '19 at 3:32
• You could start from $z=x+iy$ and apply the chain rule. Also see this and this. – Axion004 Sep 27 '19 at 3:46
• The title (“how?”) doesn't match the actual question (“why?”). – Hans Lundmark Sep 27 '19 at 7:36

The idea is that $$\frac{\partial}{\partial z}$$ and $$\frac{\partial}{\partial \bar{z}}$$ are meant to be the partial derivatives relative to the (complexified) coordinate system $$\{ z, \bar{z} \}$$. The thing you really want to have is

• $$\frac{\partial z}{\partial z} = 1$$
• $$\frac{\partial \bar{z}}{\partial z} = 0$$
• $$\frac{\partial z}{\partial \bar{z}} = 0$$
• $$\frac{\partial \bar{z}}{\partial \bar{z}} = 1$$

Or alternatively, on the basis differential forms,

• $$\frac{\partial }{\partial z} \mathrm{d} z= 1$$
• $$\frac{\partial }{\partial z} \mathrm{d} \bar{z} = 0$$
• $$\frac{\partial }{\partial \bar{z}} \mathrm{d} z= 0$$
• $$\frac{\partial }{\partial \bar{z}} \mathrm{d} \bar{z} = 1$$

These are basically the definitions of the partial derivatives. Then you can verify that $$\frac{1}{2} \left( \frac{\partial}{\partial x} - i \frac{\partial}{\partial y} \right)$$ satisfies the same identities that $$\frac{\partial}{\partial z}$$ is required to, so they are the same operator. If you didn't already know the answer, these are linear equations so you could do linear algebra to solve for the relation between them.

Incidentally, differentials play more nicely with algebra; you have

$$\mathrm{d}z = \mathrm{d}x + i \mathrm{d} y \qquad \qquad \mathrm{d}\bar{z} = \mathrm{d}x - i \mathrm{d} y$$

Does anyone have an idea why we should have this definition?

In your complex analysis book, you must also have the definition of $$\frac{\partial }{\partial \bar{z}} := \frac{1}{2}(\frac{\partial }{\partial x} +i \frac{\partial }{\partial y})$$

Now take a holomorphic function $$f$$ (on some open set $$U \subseteq \Bbb C$$ ) and apply, $$\frac{\partial }{\partial z}$$ and $$\frac{\partial }{\partial \bar{z}}$$ to $$f$$ . You will find that $$\frac{\partial f}{\partial \bar{z}}=0$$ (By Cauchy- Riemann equations). In fact $$\frac{\partial f}{\partial \bar{z}}=0 \iff f$$ is holomorphic .

So basically, the operator $$\frac{\partial }{\partial z}$$ stands for the holomorphic part of a harmonic function And $$\frac{\partial }{\partial \bar{z}}$$ precisely gives "how far it is from being holomorphic!"

In case of the Complex plane this seems tautology perhaps. But for the actual crux of the matter, one should look at Complex differential forms on Complex manifolds.

• Doesn't $\partial f/\partial \bar{z} = 0$ mean $f$ is holomorphic? You might have it backwards here. – eyeballfrog Sep 27 '19 at 4:57
• @eyeballfrog Thanks for pointing out the typo. – Brozovic Sep 27 '19 at 5:01
• @eyeballfrog In my answer I am pointing out that both are equivalent notions! IN case of the complex plane you might think it to be trivial but the actual motivation comes from looking at higher dimensional Complex manifolds! – Brozovic Sep 27 '19 at 5:04