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In Stein's Complex Analysis he gives:

Proposition:

if we write $F(x, y) = f(z)$, then $F$ is differentiable in the sense of real variables, and $$ \det J_F(x_0, y_0) = |f'(z_0)|^2. $$

Proof: To prove that $F$ is differentiable it suffices to observe that if $H = (h_1, h_2)$ and $h = h_1 + ih_2$, then the Cauchy-Riemann equations imply $$ J_F(x_0, y_0)(H) = \left( \frac{\partial u}{\partial x} - i\frac{\partial u}{\partial y} \right) (h_1 + ih_2) = f'(z_0)h, $$

MY QUESTION: Why is this not trivial? if we let $f = u +i v$, then we know the gradients of $u$ and $v$ exists as $f$ is holomorphic, then because $F(x,y) = (u(x,y) ,v(x,y))$ we have that $J_F$ is well defined. What am I missing ? To be clear I understand the way the calculation above works I just do not understand why this is a proof of the claim.

EDIT:

  1. As it has been explained in the accepted answer(By FShrike) this is not the same question as the one suggested by mittens. I was in particular wondering about this specific proof (the one I provided) , the context of the other question is much larger.

  2. Since the community seems to think (1) was not enough explanation, I will expand a little more. I understand that complex differentiability is different from $R^2$ differentiability, which was the question addressed in the suggested post, so this cannot possibly be a duplicate ....

Samael Manasseh
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1 Answers1

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To show $F$ is differentiable is to show more than mere existence of $J_F$. You need to show that $J_F$ actually satisfies the relation $\|F((x_0,y_0)+H)-F(x_0,y_0)-J_F(x_0,y_0)(H)\|\in o(H)$ as $H\to0$. It is well known that this is true if $F$'s partial derivatives in all coordinates exist and are continuous in all but possibly one coordinate. As $f$ is holomorphic it is in fact analytic and its real and imaginary components $u,v$ are in fact smooth functions so this is definitely true... so it is trivial in that sense. But we do require a fair bit of analysis to get there, and perhaps Stein-Stakarchi present this result before they prove "holomorphic $\implies$ analytic". You say:

... "we have that $J_F$ is well defined"

And you seem to suggest that that's good enough. It is not. I don't have the counterexamples off the top of my head but it is very well known to the student of multivariable calculus that "existence" of $J_F$, in the sense that you can write down the matrix of partial derivatives, is not strong enough to say that this matrix of partials actually serves its real purpose as a derivative.

The argument they present instead requires a bare minimum of analysis and gives you the "$o(H)$" condition almost for free.

FShrike
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