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Remmert's Theory of Complex Functions book is great for plenty of commentary on the history behind the mathematics introduced in each section.

The picture below is of the section which talks about Riemann's own derivation of the familiar Cauchy-Riemann conditions.

I have two questions:

  1. Where the items in the brackets of the numerator for equation (2) come from? That is, $\left(\frac{\partial u}{\partial x} + \frac{\partial v}{\partial x}i\right)$ and $\left(\frac{\partial v}{\partial y} - \frac{\partial u}{\partial y}i\right)$
  2. Is (1) Riemann's way of writing $\frac{df(z)}{dz}$ ?

enter image description here

I am not advanced, and don't have a university degree in mathematics, but I do understand the basic idea that asserting the derivative is the same, no matter which direction the limit is taken, leads to the Cauchy-Riemann relations.

Penelope
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    I can't answer your questions but I just got reading these two great posts about the Cuachy-Riemann equations before seeing your post... I highly recommend checking them out! They will add a lot to your "basic understanding" or the Cauchy-Riemann relations.

    https://math.stackexchange.com/questions/3818096/is-there-any-intuition-or-meaning-regarding-cauchy-riemann-equations

    https://math.stackexchange.com/questions/1026134/geometrical-interpretation-of-cauchy-riemann-equations/1035669

     –  Apr 17 '22 at 21:59

1 Answers1

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We start with

$$ \begin{align} du & = \frac{\partial u}{\partial x} dx + \frac{\partial u}{\partial y} dy \\ \\ dv & = \frac{\partial v}{\partial x} dx + \frac{\partial v}{\partial y} dy \end{align}$$

We construct $du + dv i$

$$ \begin{align} du + dvi & = \frac{\partial u}{\partial x} dx + \frac{\partial u}{\partial y} dy + \frac{\partial v}{\partial x} dx i + \frac{\partial v}{\partial y} dy i \\ \\ & = \left ( \frac{\partial u}{\partial x} + \frac{\partial v}{\partial x} i \right)dx + \left (\frac{\partial u}{\partial y}\frac{1}{i} + \frac{\partial v}{\partial y} \right) dyi \\ \\ & = \left ( \frac{\partial u}{\partial x} + \frac{\partial v}{\partial x} i \right )dx + \left (\frac{\partial v}{\partial y} - \frac{\partial u}{\partial y}i \right) dyi \end{align}$$

However I am not sure about the basis of the original "start with" equations.

Penelope
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    The basis of the equation $du=\frac{\partial u}{\partial x} dx + \frac{\partial u}{\partial y} dy$ is the chain rule. Write $u(x(t),y(t))$ and differentiate this w.r.t. $t$ using chain rule. Then "delete" $dt$ on both sides. – Kurt G. Apr 19 '22 at 12:55
  • Thank s@KurtG. that is helpful. Whilst I'm comfortable with "cancel dt" on both sides .. is there a theoretical rigour for doing so? Perhaps a link I can follow and read more? – Penelope Apr 19 '22 at 19:51
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    The question is how to rigorously interpret $du=\frac{\partial u}{\partial x} dx + \frac{\partial u}{\partial y} dy$ . I know of two ways of doing this: 1. treat $du,dx,dy$ as differential forms. 2. work in non standard analysis. The simplest and stiill fully rigorous approach is to do not cancel $dt$ and write it just as the good old chain rule $\frac{du(x(t),y(t))}{dt}=\frac{\partial u}{\partial x} \frac{dx}{dt} + \frac{\partial u}{\partial y} \frac{dy}{dt}$. – Kurt G. Apr 20 '22 at 08:18