Cauchy-Riemann

Friday 28 October 2005 at 2:49 pm | In Articles | 1 Comment

The Cauchy-Riemann equations are one of the first results one comes across in Complex Analysis. A poster on S.O.S. Mathematics Cyberboard has pointed that that proofs like that at Cauchy-Riemann equations tend to take it for granted that if f(x+iy)=u+iv is analytic then the partial derivatives of u and v exist. Thus the proof at Cauchy-Riemann equations says

\displaystyle f^{\prime}(z)=\lim_{h\rightarrow 0}{\left[\frac{u(x+h,y)-u(x,y)}{h}+i\frac{v(x+h,y)-v(x,y)}{h}\right]} and then deduces that \displaystyle f^{\prime}(z)=\frac{\partial u}{\partial x} + i\frac{\partial v}{\partial x}

Looking at various textbooks this omission seems to the norm. Even Ahlfors Complex Analysis says: We remark that the existence of the … partial derivatives … is implied by the existence of f^{\prime}(z)

One excellent book A First Course in Complex Functions by G.J.O. Jameson does give a proper proof of this result. It defines differentiability for u:A \to \mathbb{R}^2 (where A is a subset of \mathbb{R}^2) at a point (a,b) in the interior of A if there exists real numbers \lambda,\mu such that, given \epsilon>0, there exists \delta>0 such that, for all real h,k with \sqrt{h^2+k^2}< \delta, |u(a+h,b+k)-u(a,b)-(\lambda h + \mu k)|\le\epsilon \sqrt{h^2+k^2}

Putting k=0 shows that \lambda = \dfrac{\partial u}{\partial x}; similarly \mu = \dfrac{\partial u}{\partial y}

If f^{\prime}(a+ib)=\lambda +i\mu then, given \epsilon>0, there exists \delta>0 such that for all real h,k with |h+ik|<\delta

|f((a+h)-i(b+k))-f(a+ib)-(\lambda + i\mu)(h+ik)|\le \epsilon|h+ik|

and taking real parts

|u(a+h,b+k)-u(a,b)-(\lambda h - \mu k)|\le\epsilon \sqrt{h^2+k^2} from which it follows that \dfrac{\partial u}{\partial x} and \dfrac{\partial u}{\partial y} exist. Taking imaginary parts gives the other 2 partial derivatives.

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  1. I think that both the poster at SOS and you missed the point here. If f^{\prime}(z) exists than it follows easily that the partials exist by the fact that if one restricts oneselves to real h, than the existence of \displaystyle f^{\prime}(z)=\lim_{h\rightarrow 0}{\left[\frac{u(x+h,y)-u(x,y)}{h}+i\frac{v(x+h,y)-v(x,y)}{h}\right]}
implies that the real part of the righthand side has as limit the real part of f^{\prime}(z) (and thus the partial of u wrt x exists). This follows through the inequality |Re z -Re w|\le |z-w|. Similarly for the imaginary part.

    Comment by Anton R. Schep — Saturday 12 November 2005 2:22 am #

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