PlanetMath: proof of the Cauchy-Riemann equations

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proof of the Cauchy-Riemann equations

(Proof)

Existence of complex derivative implies the Cauchy-Riemann equations.

Suppose that the complex derivative

$\displaystyle f'(z) = \lim_{\zeta\rightarrow 0} \frac{f(z+\zeta)-f(z)}{\zeta}$

(1)

exists for some $z\in \mathbb{C}$ . This means that for all $\epsilon>0$ , there exists a $\rho>0$ , such that for all complex $\zeta$ with $\vert \zeta\vert<\rho$ , we have

$\displaystyle \left\vert f'(z) - \frac{f(z+\zeta)-f(z)}{\zeta} \right \vert<\epsilon.$

Henceforth, set

$\displaystyle f=u+iv,\quad z=x+iy.$

If $\zeta$ is real, then the above limit reduces to a partial derivative in

, i.e.

$\displaystyle f'(z) = \frac{\partial f}{\partial x} = \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x},$

Taking the limit with an imaginary $\zeta$ we deduce that

$\displaystyle f'(z) = -i\frac{\partial f}{\partial y} = -i \frac{\partial u}{\partial y} + \frac{\partial v}{\partial y}.$

Therefore

$\displaystyle \frac{\partial f}{\partial x} = -i\frac{\partial f}{\partial y},$

and breaking this relation up into its real and imaginary parts gives the Cauchy-Riemann equations.

The Cauchy-Riemann equations imply the existence of a complex derivative.

Suppose that the Cauchy-Riemann equations

$\displaystyle \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y},\quad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x},$

hold for a fixed $(x,y)\in\mathbb{R}^2$ , and that all the partial derivatives are continuous at

as well. The continuity implies that all directional derivatives exist as well. In other words, for $\xi,\eta\in\mathbb{R}$ and $\rho=\sqrt{\xi^2+\eta^2}$ we have

$\displaystyle \frac{ u(x+\xi,y+\eta) - u(x,y) - (\xi \frac{\partial u}{\partial... ...c{\partial u}{\partial y})}{\rho} \rightarrow 0,\;\mbox{as } \rho\rightarrow 0,$

with a similar relation holding for

. Combining the two scalar relations into a vector relation we obtain

$\displaystyle \rho^{-1} \left\Vert \begin{pmatrix} u(x+\xi,y+\eta) \\ v(x+\xi,y... ...d{pmatrix}\begin{pmatrix} \xi \\ \eta \end{pmatrix}\right\Vert \rightarrow 0,\;$ as $\displaystyle \rho\rightarrow 0.$

Note that the Cauchy-Riemann equations imply that the matrix-vector product above is equivalent to the product of two complex numbers, namely

$\displaystyle \left(\frac{\partial u}{\partial x} +i\frac{\partial v}{\partial x}\right)(\xi+i\eta).$

Setting

$\displaystyle f(z)$	$\displaystyle =$	$\displaystyle u(x,y)+i v(x,y),$
$\displaystyle f'(z)$	$\displaystyle =$	$\displaystyle \frac{\partial u}{\partial x} + i\frac{\partial v}{\partial x}$
$\displaystyle \zeta$	$\displaystyle =$	$\displaystyle \xi+i\eta$