Question

1 Prove that a necessary condition that

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

be analytic in a region R is that the Cauchy-Riemann equations

∂u

∂y =

∂v

∂y ,

∂u

∂y = −

∂v

∂x

are satisfied in a region R where it is supposed that these partial derivatives are

continuous in R

Answer 1

Let’s suppose that f(z) is differentiable in some region A and f(z) = f(x + iy) = u(x, y) + iv(x, y). We’ll compute f'(z) by approaching z first from the horizontal direction and then from the vertical direction. We’ll use the formula

$f'(z) = \displaystyle{{\lim}_{ ∆z→0}}\frac{ f(z + ∆z) − f(z)}{ ∆z }$

where  ∆z = ∆x + i∆y

Horizontal direction: ∆y = 0, ∆z = ∆x

$f'(z) = \displaystyle{{\lim}_{ ∆z→0}}\frac{ f(z + ∆z) − f(z)}{ ∆z } =f'(z) = \displaystyle{{\lim}_{ ∆z→0}}\frac{ f(x + ∆x+iy) − f(x+iy)}{ ∆x } =$

$= \displaystyle{{\lim}_{ ∆z→0}}\frac{ u(x + ∆x,y) +iv(x + ∆x,y)-(u(x,y)+iv(x,y))}{ ∆x } =$

$=\frac{\partial u}{\partial x}(x,y)+i\frac{\partial v}{\partial x}(x,y)$

Vertical direction: ∆x = 0, ∆z = i∆y

$f'(z) = \displaystyle{{\lim}_{ ∆z→0}}\frac{ f(z + ∆z) − f(z)}{ ∆z } =$

$=\displaystyle{{\lim}_{ ∆z→0}}\frac{ u(x,y + ∆y) +iv(x,y + ∆y)-(u(x,y)+iv(x,y))}{ i∆y } =$

$=\frac{1}{i}\frac{\partial u}{\partial y}(x,y)+\frac{\partial v}{\partial y}(x,y)=\frac{\partial v}{\partial y}(x,y)-i\frac{\partial u}{\partial y}(x,y)$

We have found two different representations of f'(z) in terms of the partials of u and v. If put them together we have the Cauchy-Riemann equations:

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

$\implies \frac{\partial u}{\partial x}(x,y)=\frac{\partial v}{\partial y}(x,y)$ and $-\frac{\partial u}{\partial y}(x,y)=\frac{\partial v}{\partial x}(x,y)$