Introduction

The Cauchy Riemann equations are a system of two partial differential equations that combined establish a necessary and sufficient condition for a complex function that is holomorphic to be complexly differentiable with specified continuity and differentiability requirements to be complexly differentiable.

The two equations are the Cauchy Riemann equation on a pair of real-valued functions of two real variables, u(x,y) and v(x,y).

They are:

 δu/δx = δv/δy   and   δu/δy = -δv/δx

The real and imaginary components are represented by u and v, respectively.

Discussion

Derivation

The derivative of f(z) with respect to z can be written using the classical definition of the derivative.

df(z)/dz = Δz→0→0 {f(z+ Δz) – f(z)}/Δz

Now, we can also use the theory of limits of complex-valued sequences and series 

So, the formal representation of the derivative becomes f(z) in terms of its real and imaginary parts will be, 

df(z)/dz = Δx→0, Δy→0  {u(x + Δx,y + y) – u(x,y) + i(v(x +Δx,y +Δy) – v(x,y))}/(Δx + iΔy)

Where Δz = Δx + iΔy.

The limit must exist and be unique in order for the derivative to exist. As a result, it should make no difference whether we take the Δx→0 first or then Δy→0 second or vice versa. In either case, we must achieve an equivalent result.

If we start with the Δy→0, the limiting process that results are

df(z)/dz = Δx→0 {u(x + Δx,y) – u(x,y) + i(v(x +Δx,y) – v(x,y))}/Δx

After solving the partial derivative with respect to x we have,

df(z)/dz = ∂u(x,y)/∂x  +i∂v(x,y)/∂x

Similarly, we have to take the limΔx→0 then we have.

df(z)/dz = Δy→0  {u(x + Δy) – u(x,y) + i(v(x+Δy) – v(x,y))}/ iΔy

df(z)/dz =  ∂u(x,y)/∂y  – i∂v(x,y)/∂y

From the above two derivations, we got two ways to represent df(z)/dz i.e.

∂u(x,y)/∂x = i∂v(x,y)/x   and   ∂u(x,y)/∂y = – i∂v(x,y)/∂y

Cauchy Riemann examples

Solution:

It has been known 

ea = ex+iy =ex cos(y) + iex sin(y).

now,

u(x,y) = ex cos(y)

hence,

v(x,y) =ex sin(y)

Now after solving the partial differential equation, we have 

ux= excos(y) ,

ux= −exsin(y) ,

vx= exsin(y) ,

uy= excos(y) 

As we can see from the above equations ux= ux and uy = –vx , hence the Cauchy Riemann equation are satisfied and ea is differentiable.

 Also, 

dez/dz=ux+ivx=excos(y)+iexsin(y)=ez.

Try to solve this by yourself rather than using the Cauchy Riemann equations calculator.

1. By using the Cauchy-Riemann equations, prove that   f(z) = z̄ cannot be differentiated.

Solution:

We know that, 

f(x+iy)=x-iyf(x+iy)=x-iy.

So, 

u(x,y)=x,v(x,y)=-y

Now after taking the partial derivative of the above equation

We have,

ux=1, uy= 0, vx=0, vx=-1

As we can see above ux uy Hence the Cauchy Riemann equations are not satisfied and therefore f is not differentiable. You can verify the result from any Cauchy Riemann equations calculator.

Historical remarks 

In 1752, Jean Le Rond d’Alembert published the original equation system that we use today. He explained the equation as follows: for any two orthogonal directions s and n with the same mutual orientation as the x and y axes, in the form:

𝜕𝑢/𝜕x = 𝜕𝑣/𝜕y, 𝜕𝑢/𝜕y = − 𝜕𝑣/𝜕x

After several years, Leonhard Rular related this system to the analytic functions in 1797.

Then, in 1914, Cauchy combined both of Alembert’s and Rular’s investigations to develop a theory of functions. Then, in 1851, Riemann’s dissertation on function theory was published.

Application of Cauchy Riemann examples in Laplace’s equation with Complex variables

Let’s have a look on the LaPlace equation in 2D, using the Cartesian coordinates:

2f/ x2 + 2f/y2 = 0.

As we can see the above equation has no real characteristics as its discriminant is negative (B2 − 4𝐴𝐶 = −4). However, if we overlook this technicality and allow for a complicated change of variables, we can profit from the same solution structure as the wave equation.

Let’s introduce a new variable “𝜂”

And,

= x + iy ; x = ( + )/ 2 ; = x – iy ; 

𝑦 = 𝜂 − 𝜉 2𝑖. 

After solving using the chain rule,

We have,

/∂x = /∂η + /∂ξ ; /∂y = i ( /∂η – /∂ξ )

Hence the  PDE will be , 

4 𝜕/𝜕𝜂 𝜕𝑓/𝜕𝜉 = 0

After solving this we get,

𝑓 = 𝑝(𝜂) + 𝑞(𝜉) = 𝑝(𝑥 + 𝑖𝑦) + 𝑞(𝑥 − 𝑖𝑦).

Assuming we desired a real solution to the original (real) PDE, p and q are differentiable complex functions.

We also have the requirement that the sum of the two functions has no imaginary parts.

This can be expressed in more standard notation:

Suppose we use the (x,y) plane to represent the complex plane in a regular manner,

Now, we introduce the complex variable “z”

And,

𝑧 = 𝑥 + 𝑖𝑦.                                       

So, the complex conjugate of z will be,

z̅ = x – iy

And hence from the above two equations, we can conclude that:

𝑝(𝑧) + 𝑞(𝑧̅).

Conclusion 

By now, it can be concluded that the idea has been clarified regarding Cauchy Riemann Equation. 

δu/δx = δv/δy   and   δu/δy = -δv/δx  where the real and imaginary components are represented by u and v, respectively.