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

we can therefore rewrite the above limit relation as
$\displaystyle \left\vert \frac{ f(z+\zeta)-f(z) - f'(z)\zeta}{ \zeta}\right\vert \rightarrow 0,\;$as $\displaystyle \rho\rightarrow 0,$
which is the complex limit definition of $ f'(z)$ shown in (1).



"proof of the Cauchy-Riemann equations" is owned by rmilson.
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Also defines:  complex derivative

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Cross-references: complex numbers, equivalent, product, vector, scalar, similar, directional derivatives, implies, continuous at, fixed, equations, Cauchy-Riemann equations, imaginary parts, relation, imaginary, partial derivative, limit, real, complex, derivative
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This is version 3 of proof of the Cauchy-Riemann equations, born on 2002-08-10, modified 2005-03-11.
Object id is 3282, canonical name is ProofOfTheCauchyRiemannEquations.
Accessed 11493 times total.

Classification:
AMS MSC30E99 (Functions of a complex variable :: Miscellaneous topics of analysis in the complex domain :: Miscellaneous)

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