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I was reading this post in order to grasp what holomorphic mean. Did I understood correctly that the analog of a holomorphic function (if we consider $\mathbb{R}^2$ as a $\mathbb{R}$-vector space) is a function with zero divergence and zero curl. Is this by doing the identification :

\begin{align}\{\text{divergence free, curl free functions in }\mathbb{R}^2\}&\to \{\text{holomorphic functions in }\mathbb{C}\}\\ (u,v)&\mapsto u-iv \end{align} and the map is well defined because of the Cauchy-Riemann relations (the maps is clearly bijective)?

Maybe this is a silly question but why do we study holomorphic functions instead of studying divergence and curl free maps of $\mathbb{R}^2$?

Arthur
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edamondo
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  • Holomorphic functions are easy to find. For example, $f(z) = z^2$ is obviously holomorphic. This gives rise to the vector field $(x^2-y^2, 2xy)$, which is much less obviously divergence/curl free (I claim). There are many other reasons. – preferred_anon Jun 28 '23 at 09:07
  • @preferred_anon That vector field is not divergence free: $\partial_x(x^2-y^2)+\partial_y(2xy)=2x+2x\not=0,,$ and it is not curl free: $\partial_x(2xy)-\partial_y(x^2-y^2)=2y+2y\not=0,.$ The socalled Harmonic vector field that is both, divergence and curl free is obtained from the complex conjugate of the holomorphic function $f,.$ – Kurt G. Jun 28 '23 at 09:29
  • Not really anything to do with the subject @KurtG., but isn't the first picture in the "Cauchy–Riemann equations" wiki article wrong? – edamondo Jun 28 '23 at 09:43
  • Happy to get to the bottom of this. Which picture exactly? Everything rests on the Cauchy-Riemann equations. The Wikipedia article writes them correctly as $$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y},,\quad \frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$$ *(1a) (1b)*. – Kurt G. Jun 28 '23 at 13:46
  • @KurtG. The picture with the caption "A visual depiction of a vector X in a domain being multiplied by a complex number z, then mapped by f, versus being mapped by f then being multiplied by $z$ afterwards." in the lower right part, shouldn't it be $df(X)$ and $zdf(X)$ instead of $df(zX)$ and $zdf(zX)$ – edamondo Jun 28 '23 at 13:56
  • I think it should be $df(zX)$ and $zdf(X),.$ Does not matter. They could have done a lot better. I recommend the following exercise: from the CR equations you can deduce that the Jacobian $df$ of $f$ is $u_x^2+u_y^2$ times an orthogonal matrix with determinant one. Now use everything you have learned from linear algebra about such matrices: what do they preserve? – Kurt G. Jun 28 '23 at 15:00
  • @KurtG. they preserve angles and distances. But how can the jacobian matrix be expressed as the product of the laplacian and an orthogonal matrix? – edamondo Jun 28 '23 at 15:36
  • Exploit the Cauchy-Riemann equations. $u_x^2+u_y^2$ is not the Laplacian $u_x,u_y$ are only first partial derivatives. Also : that Jacobian only preserves angles, not distances. – Kurt G. Jun 28 '23 at 16:34

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