Cauchy-Riemann Equations.

Recall the definition of the derivative of a function $f$ at a point $z_0$ :

$$f'(z_0) = \underset{\Delta z \rightarrow 0}{\text{lim}} \frac {f(z_0 + \Delta z) - f(z_0)}{\Delta z}$$

Denote: $z=x+iy$ , $z_0=x_0+iy_0$ , $\Delta z = \Delta x + i \Delta y$ and $f(z)=u(x,y)+i\; v(x,y)$ , where $x$ ,$y$ ,$x_0$ ,$y_0$ , $\Delta x$ ,$\Delta y$ , $u(x,y)$ and $v(x,y)$ are real.

Figure 1: Cauchy-Riemann.

Suppose that $\Delta y=0$ ; thus we have:

$$f'(z_0) = \underset{\Delta x \rightarrow 0}{\text{lim}} \frac {[u(x_0+\Delta x, y_0)+i v((x_0+\Delta x, y_0)]-[u(x_0, y_0)+i v((x_0, y_0)]}{\Delta x}$$

$$\quad =\underset{\Delta x \rightarrow 0}{\text{lim}} \frac {u(x_0+\Delt... ...ightarrow 0}{\text{lim}} \frac {v((x_0+\Delta x, y_0) - v((x_0, y_0)}{\Delta x}$$

$$\quad = u_x(x_0,y_0)+i v_x(x_0,y_0)$$

Suppose that $\Delta x=0$ ; thus we have:

$$f'(z_0) = \underset{\Delta y \rightarrow 0}{\text{lim}} \frac {[u(x_0, y_0 + \Delta y)+i v((x_0, y_0+ \Delta y)]-[u(x_0, y_0)+i v((x_0, y_0)]}{i \Delta y}$$

$$\quad = \underset{\Delta y \rightarrow 0}{\text{lim}} \frac {u(x_0, y_0... ...tarrow 0}{\text{lim}} \frac {v((x_0, y_0+ \Delta y) - v((x_0, y_0)}{i \Delta y}$$

$$\quad = - i v_y(x_0,y_0) + v_y(x_0,y_0)$$

Hence we have the so-called Cauchy-Riemann Equations:

$$\begin{cases} \frac{\partial u}{\partial x}= \frac {\partial... ...l v}{\partial x} = - \frac {\partial u}{\partial y} \end{cases}$$

which can be wriiten in the following form, with a notation frequently used in Calculus:

$$\begin{cases} u_x=v_y \ u_y=-v_x \end{cases}$$

Theorem 3.3.1   If $f(z)=u(x,y)+iv(x,y)$ is derivable at $z_0=x_0+iy_0$ , then $u$ and $v$ verify the Cauchy-Riemann Equations at $(x_0,y_0)$ .

Example 3.3.2   Let $f(z)=z^2$ . As a polynomial function, $f$ is derivable over the whole of $\mathbb{C}$ .

Let us check the Cauchy-Riemann equations. Denote $z=x+iy, \; x,y \in \mathbb{R}$ . Then we have:

$$f(z)=(x+iy)^2=\underbrace{x^2-y^2}_{=u(x,y)}+i \; \underbrace{2xy}_{=v(x,y)}.$$

It follows that:

$$\begin{cases}u_x=2x=v_y \ u_y=-2y=-2v_x \end{cases}$$

at every point in the plane, i.e. Cauchy-Riemann equations hold everywhere.

Example 3.3.3   Let $f(z)=\vert z\vert^2$ . If $z=x+iy$ , where $x,y$ are real numbers, then:

$$f(z)=x^2+y^2=\underbrace{x^2+y^2}_{u(x,y)}+ i \cdot \underbrace{0}_{v(x,y)}$$

Let us check at which points the Cauchy-Riemann equations are verified. We have: $u_x=2x$ , $u_y=2y$ , $v_x=0$ and $v_y=0$ .Cauchy-Riemann equations are verified if, and only if, $$\begin{cases}2x=0 \ 2y=0\end{cases}$$ , i.e. $x=y=0$ . The only point where $f$ can be differentiable is the origin.

There is a kind of inverse theorem:

Theorem 3.3.4   If $f(z)=u(x,y)+iv(x,y)$ verifies the Cauchy-Riemann Formulas at $z_0$ and if the partial derivatives of $u$ and $v$ are continuous at $(x_0,y_0)$ , the $f$ is derivable at $z_0$ and $f'(z_0)=u_x(x_0,y_0)+iv_x(x_0,y_0)$ .

Example 3.3.5   Let $f(z)=z^2$ ; the function $f$ is derivable at any point and $f'(z)=2z$ .

If $z=x+iy$ , then $f(z)=(x+iy)^2=\underbrace{x^2-y^2}_{u(x,y)} + i \underbrace{2xy}_{v(x,y)}$ . Then: $u_x=2x$ , $u_y=-2y$ , $v_x=2y$ and $v_y=2y$ . These partial derivatives verify the C-R equations.

By that way, we have a new proof of the differentiability of $f$ at every point.

Example 3.3.6   Let $f(z)=z\vert z\vert^2$ , for any $z \in \mathbb{C}$ . We work as in the previous examples:

$$f(z)=(x+iy)(x^2+y^2) = \underbrace{(x^3+xy^2)}_{=u(x,y)}+i \underbrace{(x^2y+y^3)}_{=v(x,y)}$$

We compute the first partial derivatives:

$$\begin{cases}u_x=3x^2+y^2 \ u_y = 2xy \end{cases} \qquad \text{and} \qquad \begin{cases}v_x=2xy \ v_y = x^2+3y^2 \end{cases}$$

We solve Cauchy-Riemann equations:

$$\begin{cases}3x^2+y^2=x^2+3y^2 \ 2xy=-2xy \end{cases} \Longleftrightarrow [ \; x=0 \quad \text{or} \quad y=o \; ]$$

The subset of the plane where $f$ can be differentiable is the union of the two coordinate axes. As the first partial derivatives of $u$ and $v$ are continuous at every point in the plane, $f$ is differentiable at every point on one of the coordinate axes.

Cauchy-Riemann equations in polar form:

Instead of $f(z)=u(x,y)+iv(x,y)$ , write $f(z)=u(r,\theta)+iv(r,\theta)$ , where $z=x+iy=r(\cos \theta+i \sin \theta)$ .

$$\begin{cases} \frac {\partial u}{\partial r} = \frac {1}{r} ... ...ac {1}{r} \cdot \frac {\partial u}{\partial \theta} \end{cases}$$