Question on derivative of conjugate

[I like Serena]

In particular, in this, Riemann's original interpretation, and the interpretation now used in the theory of complex and almost complex manifolds, everything jackmell said originally is correct, up until he began to doubt himself.

Can you elaborate on how everything what jackmell said was correct?

 

[mathwonk]

i have already explained it above but will try again. in particular jackmell was correct in saying if f(z) = zbar, then ∂f/∂z = 0 and ∂f/∂zbar = 1.

(I do not agree with him that one cannot write f(z) = zbar, but in a sense his choice is more suggestive.)

This is by definition of the operators ∂/∂z and ∂/∂ zbar.

I.e. if f is any smooth map, then the differential of f is defined as df = ∂f/∂x dx + ∂f/∂y dy.

This is the differential of a smooth map from C to C, i.e. from R^2 to R^2. Thus at each point, the differential is an R- linear map from C to C.

But Riemann proposed to study smooth maps whose differential was in fact complex linear.

To determine which smooth maps these were, he decomposed the differential into its complex linear and complex conjugate linear parts.Now the real linear maps from C to C form a two dimensional complex vector space, with complex basis the functions x and y. The coefficients of df in that basis are ∂f/∂x and ∂f/∂y.

This is equivalent to saying df = ∂f/∂x dx + ∂f/∂y dy.

But there is another complex basis for the real linear functions from C to C, namely z and zbar.This reflects the fact that the real linear functions decompose into a direct sum of the subspaces of complex linear and conjugate linear maps. I.e. z is a complex basis of the complex linear maps, and zbar is a complex basis of the conjugate linear maps.

The coefficients of df in this basis are called ∂f/∂z and ∂f/∂zbar.

It is already obvious that if f(z) = zbar, then ∂f/∂z = 0 and df/∂zbar = 1.

I.e. zbar is real linear so equals its own best linear approximation at each point.

But it also is conjugate linear, so at each point the differential of zbar equals zbari.e. it equals 0.z + 1.zbar. these coefficients are the ∂/∂z and ∂/∂zbar derivatives,

which thus equal 0 and 1.
This is also equivalent to saying

that df = ∂f/∂z dz + ∂f/∂zbar dzbar. We can thus just calculate ∂f/∂ and ∂f/∂zbar

mechanically. I.e.

Since z + zbar = 2x and z - zbar =2iy, we can change coordinates in a basis expansion by

substitution. I.e. if a linear map equals ax + by, then substituting x = (1/2)(z+zbar),

and y = (1/2i)(z-zbar), we get ax + by = a(1/2)(z+zbar) + b(1/2i)(z-zbar)Now if a = ∂f/∂x and b = ∂f/∂y, then for f(z) = zbar, we have a = 1, b = -i.

Hence we get df = (1/2)(z+zbar) -i (1/2i)(z-zbar)

= (1/2)(z+zbar) - (1/2)(z-zbar) = 0.z + 1.zbar = ∂f/∂z dz + ∂f/∂zbar dzbar.

So ∂f/∂z =0 , and ∂f/∂zbar = 1.Or you can just use micromass' equations for the operators

∂/∂z = (1/2)(∂/∂x - i∂/∂y), and ∂/∂zbar = (1/2)(∂/∂x + i∂/∂y).

since for f(z) = zbar, we have again ∂f/∂x = 1 and ∂f/∂zbar = -i, substituting in gives

us ∂f/∂z = (1/2)(1-1) = 0, and ∂f/∂zbar = (1/2)(1 + 1) = 1.

 

[mathwonk]

Moreover if you have a complex function expressed by formulas in terms of x and y, you can substitute as above, x= (1/2)(z+zbar) and y = (1/2i)(z-zbar) and get an expression in z and zbar. This works for explicit formulas at least, which will presumably be harmonic functions. Then taking the ∂/∂z and ∂/∂zbar derivatives would amount to differentiating in the same way as in taking partials wrt these separate variable.

But the discussion above shows that even when you cannot separate the function into functions of z and zbar separately, you can still separate the derivative into complex linear and conjugate linear parts.

some people still like to write the function as f(z,zbar) for suggestive reasons. So actually he was right about that too, in this sense.

am i making any sense?

 

[mathwonk]

wow! I didn't realize this before, but you CAN always treat this as differentiating in the z and zbar directions.

E.g. suppose you want to differentiate sin(x) wrt z and zbar.

Just rewrite sin(x) = sin([z+zbar]/2) and take ∂/∂z and ∂/∂zbar in the same way as you take any partials. I.e. to take ∂/∂z, consider zbar as constant. Of course you have to start from the facts that ∂zbar/∂z = 0 = ∂z/∂zbar, and ∂z/∂z = 1 = ∂zbar/∂zbar.

e.g. ∂/∂z sin(x) = ∂/∂z sin([z+zbar]/2) = cos([z+zbar]/2). (1/2) = (1/2)cos(x).

and ∂/∂zbar sin(x) = ∂/∂zbar sin([z+zbar]/2) = cos([z+zbar]/2). (1/2) = (1/2)cos(x).

Thus if f(z,zbar) = sin(x) = sin([z+zbar]/2), then df = ∂f/∂z dz + ∂f/∂zbar dzbar

= (1/2)cos(x) dz + (1/2)cos(x). dzbar

= (1/2)cos(x) (dx + idy) + (1/2)cos(x) (dx-idy) = cos(x) dx.Yes! It seems you can always substitute x = (1/2)(z+zbar) and y = (1/2i)(z-zbar),

and consider any complex function f(x,y) as a function f(z,zbar),

and then one can take the ∂/∂z and ∂/dzbar derivatives exactly as when taking partials.

So now I even think jackmell was right about wanting to write zbar as f(z,zbar) = zbar.he was wrong to doubt himself and say that f would be analytic if the derivative were zero everywhere. Rather since it was the ∂/∂zbar derivative that was zero, it means the function is antiholomorphic.

 

[mathwonk]

Example:

f(x,y) = e^x.cos(y) + i.e^x.sin(y).

then ∂f/∂x = e^x.cos(y) +i.e^x.sin(y).

∂f/∂y = -e^x.sin(y) + i.e^x.cos(y).

then ∂f/∂z = (1/2)(∂f/∂x –i∂f/∂y) = e^x.cos(y) +i.e^x.sin(y) = ∂f/∂x.

and ∂f/∂zbar = (1/2)(∂f/∂x + i∂f/∂y) = 0.

Thus f is independent of zbar and depends only on z, hence f is holomorphic.That implies that f can be written as an analytic function of z alone. To see that, substitute x = (1/2)(z+zbar), and y = (1/2i)(z-zbar),

And get f(z,zbar) = e^((z+zbar)/2).cos(y) + i.e^x.sin(y). I lost focus. I don’t feel like substituting all this in and still not gettong anywhere.

So I’ll just say that f(z,zbar) = e^z, independent of zbar.

 

[A. Bahat]

While on the topic of dz and dz-bar, it's nice to note that using differential forms in complex analysis brings out the deep connections with multivariable calculus. (After all, C is just R^2 with multiplication.)

First, note that if f is holomorphic in some region, then f dz is closed in that region. Upon applying Stokes' theorem, Cauchy's integral theorem follows as a trivial corollary. (The integral of f dz over the boundary equals the integral of d(f dz)=0 over the interior.) Once this connection is established, it gives the theorem a wonderful physical interpretation: a conservative force does no work over a closed path.

If f is instead meromorphic in a region, then the residue theorem follows similarly: this time the integral of f dz over a loop gives the sum of the integrals of small loops around the poles, which give the residues (up to winding numbers and a multiplicative constant).

 

[mathwonk]

some of the confusion is the distinction between ∂/∂z and d/dz. As defined in usual complex analysis courses, df/dz does not exist unless f is holomorphic, but ∂f/∂z exists for every smooth f.

Hence if f(z) = zbar, then df/dz does not exist, but ∂f/∂z = 0.

 

[A. Bahat]

Moreover, now the Cauchy-Riemann equations take the suggestive form ∂f/∂z-bar=0, i.e. f is only a function of z.

 

[mathwonk]

this is what i tried to say at the bottom of post 29.