Cauchy-Riemann equations

In complex analysis, the Cauchy-Riemann equations are the partial differential equations that hold if $f(z)=u+iv$ is complex differentiable at a point $z=x_0+i{y}_0$:

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

These have broad implications in other parts of complex analysis, and are often used to simplify complex expressions or within proofs.

In particular, if a function satisfies the Cauchy-Riemann equations in its neighbourhood and the function and its partials are continuous, it is differentiable. If it isn’t differentiable, then it doesn’t satisfy the equations. We can also express the derivative as:

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

Proof

Suppose the definition for differentiability holds at a given point $z_0=x_0+i{y}_0$. Let’s first compute the limit as we approach $z_0$ along the horizontal lie $z=x+i{y}_0$.

$$f^{\prime }(x+i{y}_0)=\underset{x\to 0}{\lim }\frac{f(x+i{y}_0)-f(x_0+i{y}_0)}{x+i{y}_0-(x_0+i{y}_0)}$$

After some complicated manipulations:

$$f^{\prime }(x+i{y}_0)=\underset{x\to x_0}{\lim }\frac{u(x,y)+iv(x,y_0)-u(x_0,y_0)+iv(x_0,y_0)}{x-x_0}$$ $$f^{\prime }(x+i{y}_0)=\frac{\partial u}{\partial x}(x_0,y_0)+i\frac{\partial v}{\partial x}(x_0,y_0)$$

If differentiability holds, we must also have the same limit as we approach $z=x_0+i{y}_0$ along the vertical line $x=x_0$.

$$f^{\prime }(x+i{y}_0)=\frac{\partial v}{\partial y}(x_0,y_0)-i\frac{\partial u}{\partial y}(x_0,y_0)$$

From above, we must have the same limit, so we can equate the real and imaginary parts to get the Cauchy-Riemann equations as stated.