proof of the Cauchy-Riemann equations

Existence of complex derivative implies the Cauchy-Riemannequations.

Suppose that the complexderivative

$$f^{\prime }(z)=\underset{\zeta \to 0}{lim}\frac{f(z+\zeta )-f(z)}{\zeta }$$(1)

exists for some $z\in ℂ$.This means that for all $ϵ>0$, there exists a $\rho >0$, such thatfor all complex $\zeta$ with, we have

Henceforth, set

$$f=u+iv,z=x+iy.$$

If $\zeta$ isreal, then the above limit reduces to a partial derivative in $x$, i.e.

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

Taking the limit with animaginary $\zeta$ we deduce that

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

Therefore

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

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

The Cauchy-Riemannequations imply the existence of a complex derivative.

Suppose that the Cauchy-Riemannequations

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

hold for a fixed $(x,y)\in {ℝ}^2$,and that all thepartial derivatives are continuous at $(x,y)$ as well. The continuityimplies that all directional derivatives exist as well. Inother words, for $\xi ,\eta \in ℝ$ and $\rho =\sqrt{{\xi }^2+{\eta }^2}$we have

$$\frac{u(x+\xi ,y+\eta )-u(x,y)-(\xi \frac{\partial u}{\partial x}+\eta \frac{\partial u}{\partial y})}{\rho }\to 0,\text{as }\rho \to 0,$$

with a similar relation holding for $v(x,y)$. Combining the two scalarrelations into a vector relation we obtain

Note thatthe Cauchy-Riemann equations imply that the matrix-vector productabove 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

$f(z)$$=$$u(x,y)+iv(x,y),$

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

$\zeta$$=$$\xi +i\eta$

we can therefore rewrite the above limit relation as

$$\left|\frac{f(z+\zeta )-f(z)-f^{\prime }(z)\zeta }{\zeta }\right|\to 0,\text{as }\rho \to 0,$$

which is the complex limit definition of $f^{\prime }(z)$ shownin (1).