I don't use complex analysis much in my "day job", but I'm perusing the materials for personal interest. I just wanted to check with the community about my understanding of complex analytic (holomorphic) functions, and why they are "special" compared to real-valued functions.

A far as I understood it, the definition of a complex holomorphic function is a consequence of the fact that complex numbers essentially live in $\mathbb{R}^2$ but has a funny definition of algebraic multiplication $(x_1,y_1)(x_2,y_2) = (x_1 x_2 - y_1 y_2, x_1 y_2 + x_2 y_1)$ that makes it useful for defining roots of negative numbers, while at the same time still acting like real numbers when the $y$ component is zero.

Complex functions, since they can essentially be mapped to something that's $f:\mathbb R^2 \rightarrow \mathbb R^2$, have by definition derivatives that are Frechet derivatives ($2 \times 2$ linear operators in this case). However, the derivative also need to act like a single complex number, so the $2 \times 2$ linear operator needs to act like multiplication by a complex number. By the definition of complex products above, this means there are special restrictions on the matrix elements of this operator. And by extension of the fact that Frechet derivatives is just a matrix of partial derivatives, special relationships exist between the partial derivatives.

Quite a bit down the road, this essentially leads to the consequences laid out in the Cauchy Integral Theorem. Is that correct?

TLDR: the definition of complex number products, along with their $\mathbb R^2$ nature, leads to the definition of holomorphic functions and the Cauchy Integral Theorem. Correct?

  • 6
    $\begingroup$ Basically correct, if stated a bit peculiarly. Your "special restrictions on the matrix elements of this operator" are called the "Cauchy-Riemann equations". $\endgroup$ – GEdgar Oct 14 '11 at 0:15
  • $\begingroup$ If I am linking things correctly, the Cauchy-Reimann basically requires the Frechet derivative to be Hermitian self-adjoint (and thus behave "like a scalar"). Which nothin came first, CR or self-adjointness? $\endgroup$ – Tim Lin Oct 14 '11 at 0:24
  • $\begingroup$ No, it is not Hermitian self-adjoint. Probably you need to spend some time with a complex analysis textbook and learn the more conventional terminology. After that, come back and see what it looks like as a "special" $\mathbb R^2$ function. $\endgroup$ – GEdgar Oct 14 '11 at 15:33

Your Answer

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

Browse other questions tagged or ask your own question.