ProofOfTheCauchyRiemannEquations 2002-08-10 06:52:37 2005-03-11 12:27:07 Proof Cauchy-Riemann equations 0 complex derivative \usepackage{amsmath} \usepackage{amsfonts} \usepackage{amssymb} \newcommand{\reals}{\mathbb{R}} \newcommand{\natnums}{\mathbb{N}} \newcommand{\cnums}{\mathbb{C}} \newcommand{\znums}{\mathbb{Z}} \newcommand{\lp}{\left(} \newcommand{\rp}{\right)} \newcommand{\lb}{\left[} \newcommand{\rb}{\right]} \newcommand{\supth}{^{\text{th}}} \newtheorem{proposition}{Proposition} \newtheorem{definition}[proposition]{Definition} \newcommand{\nl}[1]{{\PMlinkescapetext{{#1}}}} \newcommand{\pln}[2]{{\PMlinkname{{#1}}{#2}}} \paragraph{Existence of complex derivative implies the Cauchy-Riemann equations.} Suppose that the complex derivative \begin{equation} \label{eq:cder} f'(z) = \lim_{\zeta\rightarrow 0} \frac{f(z+\zeta)-f(z)}{\zeta} \end{equation} exists for some $z\in \cnums$. 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 $$\left\vert f'(z) - \frac{f(z+\zeta)-f(z)}{\zeta} \right \vert<\epsilon.$$ Henceforth, set $$f=u+iv,\quad z=x+iy.$$ If $\zeta$ is real, then the above limit reduces to a partial derivative in $x$, i.e. $$f'(z) = \frac{\partial f}{\partial x} = \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x},$$ Taking the limit with an imaginary $\zeta$ we deduce that $$f'(z) = -i\frac{\partial f}{\partial y} = -i \frac{\partial u}{\partial y} + \frac{\partial v}{\partial y}.$$ Therefore $$\frac{\partial f}{\partial x} = -i\frac{\partial f}{\partial y},$$ and breaking this relation up into its real and imaginary parts gives the Cauchy-Riemann equations. \paragraph{The Cauchy-Riemann equations imply the existence of a complex derivative.} Suppose that the Cauchy-Riemann equations $$ \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y},\quad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}, $$ hold for a fixed $(x,y)\in\reals^2$, 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 $\xi,\eta\in\reals$ and $\rho=\sqrt{\xi^2+\eta^2}$ we have $$ \frac{ u(x+\xi,y+\eta) - u(x,y) - (\xi \frac{\partial u}{\partial x } + \eta \frac{\partial u}{\partial y})}{\rho} \rightarrow 0,\;\mbox{as } \rho\rightarrow 0,$$ with a similar relation holding for $v(x,y)$. Combining the two scalar relations into a vector relation we obtain $$ \rho^{-1} \left\Vert \begin{pmatrix} u(x+\xi,y+\eta) \\ v(x+\xi,y+\eta) \end{pmatrix} - \begin{pmatrix} u(x,y) \\ v(x,y) \end{pmatrix} - \begin{pmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y}\\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}\\ \end{pmatrix} \begin{pmatrix} \xi \\ \eta \end{pmatrix} \right\Vert \rightarrow 0,\;\mbox{as } \rho\rightarrow 0.$$ Note that the Cauchy-Riemann equations imply that the matrix-vector product above is equivalent to the product of two complex numbers, namely $$\left(\frac{\partial u}{\partial x} +i\frac{\partial v}{\partial x}\right)(\xi+i\eta).$$ Setting \begin{eqnarray*} f(z) &=& u(x,y)+i v(x,y),\\ f'(z) &=& \frac{\partial u}{\partial x} + i\frac{\partial v}{\partial x}\\ \zeta &=& \xi+i\eta \end{eqnarray*} we can therefore rewrite the above limit relation as $$ \left\vert \frac{ f(z+\zeta)-f(z) - f'(z)\zeta}{ \zeta}\right\vert \rightarrow 0,\;\mbox{as } \rho\rightarrow 0,$$ which is the complex limit definition of $f'(z)$ shown in \eqref{eq:cder}.