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Revision difference : proof of the Cauchy-Riemann equations
Version 2 Version 1
\paragraph{Existence of complex derivative implies the Cauchy-Riemann \paragraph{Existence of complex derivative implies the Cauchy-Riemann
equations.} equations.}
Suppose that the complex Suppose that the complex
derivative derivative
\begin{equation} \begin{equation}
\label{eq:cder} \label{eq:cder}
f'(z) = \lim_{\zeta\rightarrow 0} \frac{f(z+\zeta)-f(z)}{\zeta} f'(z) = \lim_{\zeta\rightarrow 0} \frac{f(z+\zeta)-f(z)}{\zeta}
\end{equation} \end{equation}
exists for some $z\in \cnums$. exists for some $z\in \cnums$.
This means that for all $\epsilon>0$, there exists a $\rho>0$, such that This means that for all $\epsilon>0$, there exists a $\rho>0$, such that
for all complex $\zeta$ with for all complex $\zeta$ with
$\vert \zeta\vert<\rho$, we have $\vert \zeta\vert<\rho$, we have
$$\left\Vert f'(z) - \frac{f(z+\zeta)-f(z)}{\zeta} \right \Vert<\epsilon.$$ $$\left\Vert f'(z) - \frac{f(z+\zeta)-f(z)}{\zeta} \right \Vert<\epsilon.$$
Henceforth, set Henceforth, set
$$f=u+iv,\quad z=x+iy.$$ $$f=u+iv,\quad z=x+iy.$$
If $\zeta$ is If $\zeta$ is
real, then the above limit reduces to a partial derivative in $x$, i.e. real, then the above limit reduces to a partial derivative in $x$, i.e.
$$f'(z) = \frac{\partial f}{\partial x} = \frac{\partial u}{\partial $$f'(z) = \frac{\partial f}{\partial x} = \frac{\partial u}{\partial
x} + i \frac{\partial v}{\partial x},$$ x} + i \frac{\partial v}{\partial x},$$
Taking the limit with an Taking the limit with an
imaginary $\zeta$ we deduce that imaginary $\zeta$ we deduce that
$$f'(z) = -i\frac{\partial f}{\partial y} = -i \frac{\partial u}{\partial $$f'(z) = -i\frac{\partial f}{\partial y} = -i \frac{\partial u}{\partial
y} + \frac{\partial v}{\partial y}.$$ y} + \frac{\partial v}{\partial y}.$$
Therefore Therefore
$$\frac{\partial f}{\partial x} = -i\frac{\partial f}{\partial y},$$ $$\frac{\partial f}{\partial x} = -i\frac{\partial f}{\partial y},$$
and breaking this relation up into its real and imaginary parts gives and breaking this relation up into its real and imaginary parts gives
the Cauchy-Riemann equations. the Cauchy-Riemann equations.
\paragraph{The Cauchy-Riemann \paragraph{The Cauchy-Riemann
equations imply the existence of a complex derivative.} equations imply the existence of a complex derivative.}
Suppose that the Cauchy-Riemann Suppose that the Cauchy-Riemann
equations equations
$$
\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y},\quad \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y},\quad
\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}, \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x},
$$
hold for a fixed $(x,y)\in\reals^2$, hold for a fixed $(x,y)\in\reals^2$,
and that all the and that all the
partial derivatives are continuous at $(x,y)$ as well. The continuity partial derivatives are continuous at $(x,y)$ as well. The continuity
implies that all directional derivatives exist as well. In implies that all directional derivatives exist as well. In
other words, for $\xi,\eta\in\reals$ and $\rho=\sqrt{\xi^2+\eta^2}$ other words, for $\xi,\eta\in\reals$ and $\rho=\sqrt{\xi^2+\eta^2}$
we have we have
$$
\frac{ \frac{
u(x+\xi,y+\eta) - u(x,y) - (\xi \frac{\partial u}{\partial x } + \eta u(x+\xi,y+\eta) - u(x,y) - (\xi \frac{\partial u}{\partial x } + \eta
\frac{\partial u}{\partial y})}{\rho} \rightarrow \frac{\partial u}{\partial y})}{\rho} \rightarrow
0,\;\mbox{as } \rho\rightarrow 0,$$ 0,\;\mbox{as } \rho\rightarrow 0,$$
with a similar relation holding for $v(x,y)$. Combining the two scalar with a similar relation holding for $v(x,y)$. Combining the two scalar
relations into a vector relation we obtain relations into a vector relation we obtain
$$
\rho^{-1} \left\Vert \rho^{-1} \left\Vert
\begin{pmatrix} \begin{pmatrix}
u(x+\xi,y+\eta) \\ v(x+\xi,y+\eta) u(x+\xi,y+\eta) \\ v(x+\xi,y+\eta)
\end{pmatrix} \end{pmatrix}
-
\begin{pmatrix} \begin{pmatrix}
u(x,y) \\ v(x,y) u(x,y) \\ v(x,y)
\end{pmatrix} \end{pmatrix}
-
\begin{pmatrix} \begin{pmatrix}
\frac{\partial u}{\partial x} & \frac{\partial u}{\partial y}\\ \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y}\\
\frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}\\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}\\
\end{pmatrix} \end{pmatrix}
\begin{pmatrix} \begin{pmatrix}
\xi \\ \eta \xi \\ \eta
\end{pmatrix} \end{pmatrix}
\right\Vert \rightarrow 0,\;\mbox{as } \rho\rightarrow 0.$$ \right\Vert \rightarrow 0,\;\mbox{as } \rho\rightarrow 0.$$
Note that Note that
the Cauchy-Riemann equations imply that the matrix-vector product the Cauchy-Riemann equations imply that the matrix-vector product
above is equivalent to the product of two complex numbers, namely above is equivalent to the product of two complex numbers, namely
$$\left(\frac{\partial u}{\partial x} +i\frac{\partial v}{\partial $$\left(\frac{\partial u}{\partial x} +i\frac{\partial v}{\partial
x}\right)(\xi+i\eta).$$ x}\right)(\xi+i\eta).$$
Setting Setting
\begin{eqnarray*} \begin{eqnarray*}
f(z) &=& u(x,y)+i v(x,y),\\ f(z) &=& u(x,y)+i v(x,y),\\
f'(z) &=& \frac{\partial u}{\partial x} + i\frac{\partial v}{\partial f'(z) &=& \frac{\partial u}{\partial x} + i\frac{\partial v}{\partial
x}\\ x}\\
\zeta &=& \xi+i\eta \zeta &=& \xi+i\eta
\end{eqnarray*} \end{eqnarray*}
we can therefore rewrite the above limit relation as we can therefore rewrite the above limit relation as
$$
\left\Vert \frac{ f(z+\zeta)-f(z) - f'(z)\zeta}{ \zeta}\right\Vert \rightarrow \left\Vert \frac{ f(z+\zeta)-f(z) - f'(z)\zeta}{ \zeta}\right\Vert \rightarrow
0,\;\mbox{as } \rho\rightarrow 0,$$ 0,\;\mbox{as } \rho\rightarrow 0,$$
which is the complex limit definition of $f'(z)$ shown which is the complex limit definition of $f'(z)$ shown
in \eqref{eq:cder}. in \eqref{eq:cder}.