Cauchy-Riemann Equations: Statement, Proof, Questions

The Cauchy Riemann equation states that if a complex function f(z) is differentiable at z=z₀, then if_x(z₀) = f_y(z₀). This is one of the main tool to test whether a complex function is differentiable or not. Here we state and prove Cauchy-Riemann equations along with some solved examples as an application.

Statement of Cauchy-Riemann Equations

Let f(z) = u(x,y) + iv(x,y) be a complex function, where u and v are two real functions in variables x, y. The Cauchy-Riemann equation (C-R equation) states that if f(z) is differentiable at z₀=x₀+iy₀ if

$\boxed{\dfrac{\partial u}{\partial x}=\dfrac{\partial v}{\partial y}, \quad \dfrac{\partial u}{\partial y} = -\dfrac{\partial v}{\partial x}}$

holds at (x₀, y₀).

Proof of Cauchy-Riemann Equations

Suppose that f(z) = f(x, y) = u(x,y) + iv(x,y) is differentiable at z₀=x₀+iy₀. Then by definition, the following limit

$f'(z_0)=$ lim_h→0 $\dfrac{f(z_0+h)-f(z_0)}{h}$

exists and is independent of the path where h→0.

Let us compute the above limit along the below two paths.

Path 1: Let h=h₁+i⋅0. So h₁→0 when h→0. Note that z₀+h = (x₀+h₁)+iy₀. Now,

$f'(z_0)=$ limh→0 $\dfrac{f(x_0+h_1, y_0)-f(x_0, y_0)}{h}$ $=\dfrac{\partial f}{\partial x}(x_0, y_0)$ as h=h1.

Path 2: Let h=0+i⋅h₂. So h₂→0 when h→0. Note that z₀+h = x₀+i(y₀+h₂). Now,

$f'(z_0)=$ limh→0 $\dfrac{f(x_0, y_0+h_2)-f(x_0, y_0)}{h}$ $=\dfrac{1}{i} \dfrac{\partial f}{\partial y}(x_0, y_0)$ as h=ih2.

As the quantity $f'(z_0)$ is independent of the path along which h→0, we deduce that the values of $f'(z_0)$ along the above two paths are equal. In other words,

$\dfrac{\partial f}{\partial x}(x_0, y_0) = \dfrac{1}{i} \dfrac{\partial f}{\partial y}(x_0, y_0)$ ⇒ $i \left( \dfrac{\partial u}{\partial x} + i \dfrac{\partial u}{\partial y} \right) (x_0, y_0)$ $= \left( \dfrac{\partial v}{\partial x} + i \dfrac{\partial v}{\partial y} \right) (x_0, y_0)$ ⇒ $\left( -\dfrac{\partial u}{\partial y} + i \dfrac{\partial u}{\partial x} \right)(x_0, y_0)$ $= \left( \dfrac{\partial v}{\partial x} + i \dfrac{\partial v}{\partial y} \right) (x_0, y_0)$

Equating the real and imaginary parts, we obtain the C-R equations $\frac{\partial u}{\partial x} (x_0, y_0) = \frac{\partial v}{\partial y} (x_0, y_0)$ and $\frac{\partial u}{\partial y}(x_0, y_0) = -\frac{\partial v}{\partial x}(x_0, y_0)$.

Remark:

1. Note that since $\frac{\partial}{\partial \overline{z}}:=\frac{\partial}{\partial x} \frac{\partial x}{\partial \overline{z}} + \frac{\partial}{\partial y} \frac{\partial y}{\partial \overline{z}}$, the partial derivative of f=u+iv with respect to $\overline{z}$ is given as follows: $f_{\overline{z}}=\frac{\partial f}{\partial \overline{z}}$ $=\frac{1}{2} (\frac{\partial f}{\partial x} + i \frac{\partial f}{\partial y})$ $=\frac{1}{2} ([\frac{\partial u}{\partial x}-\frac{\partial v}{\partial y}]$ $+ i [\frac{\partial u}{\partial y}-\frac{\partial v}{\partial x}])$. Thus, by C-R equations, a necessary condition for a complex function f to be differentiable at z₀ is that $f_{\overline{z}}(z_0)=0$.
2. The converse of C-R equations is not true always.

Also Read: Complex Differentiation: Definition, Solved Problems

Analytic Function: Definition, Examples, Properties

Solved Problems

Question 1: Show that the function f(z) = Re(z) does not satisfy Cauchy-Riemann equations.

Answer:

Let z=x+iy.

Thus, f(z) = Re(z) = x. If f=u+iv then it follows that u=x, v=0.

Consider a complex number z₀. Now,

$\frac{\partial u}{\partial x}(z_0)=1$, $\frac{\partial v}{\partial y}(z_0)=0$, $\frac{\partial u}{\partial y}(z_0) =0$, $\frac{\partial v}{\partial x}(z_0)=0$.

Therefore,

$\frac{\partial u}{\partial x}(z_0) \neq \frac{\partial v}{\partial y}(z_0)$.

So the function f(z) = Re(z) does not satisfy C-R equations at z₀. As z₀ is an arbitrary point, Re(z) does not satisfy C-R equations on the whole complex plane.

Question 2: Show that f(z) = Re(z) is nowhere differentiable.

Answer:

From Q1, we see that f(z)=Re(z) does not satisfy C-R equations at any complex number. Thus, it is not differentiable at any point. In other words, the function f(z) = Re(z) is nowhere differentiable.

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