Re: Complex differentiable and multiplication

From: David C. Ullrich (ullrich_at_math.okstate.edu)
Date: 10/02/04 

Date: Sat, 02 Oct 2004 08:42:24 -0500

On Sat, 2 Oct 2004 11:37:56 +0000 (UTC), anonymous@mathforum.org
(Sergei) wrote:

>
> Could someone explain to me the meaning of this statement:
>
> (From Mathworld)
>
> Re: Complex-differentiable functions:
>
> "...Then f is complex-differentiable if its' Jacobian is
> of the form:
>
>
> [a b] (*)
> [-b a]
>
> ...."
>
> This follows from C-R, but , below is where I am not clear:
>
> That is, the derivative is given by the multiplication of
> a complex number".

The derivative they're talking about here is a real-linear map
from R^2 to R^2 given by that matrix of partial derivatives.
Say T : R^2 -> R^2 is real-linear. Then it's "given by multiplication
by a complex number" if and only if it's equivalent to a complex-
linear map from C to C (equivalent using the standard identification
of C with R^2). And T is complex-linear if and only if it commutes
with multiplication by i: T(iz) = iT(z).

Now say z = x + iy, T is defined by the matrix

  [a b]
  [c d],

note that i(x,y) = (-y,x), and verify that T commutes with
multiplcation by i if and only if c = -b and d = a.

> I understand that there is an isomorphism from C to
> GL(4,R) given by a+bi->[a b] .
> [-b a]
>
> And I know that complex multiplication implies both
> scaling and rotating, but how does the above representation
> of the Jacobian (*) relate analytic functions with
> multiplication?
> Thanks for any help.
>
>
>

************************

David C. Ullrich