cauchy riemann equation

Cauchy-Riemann Equations. Recall the definition of the derivative of a function $ f$ at a point $ z_0$ :

 

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



 

 

Denote: $ z=x+iy$ ,

$ z_0=x_0+iy_0$ ,

$ \Delta z = \Delta x + i \Delta y$ and

$ f(z)=u(x,y)+i\;<br />
v(x,y)$ , where $ x$ ,$ y$ ,$ x_0$ ,$ y_0$ , $ \Delta<br />
x$ ,$ \Delta y$ , $ u(x,y)$ and $ v(x,y)$ are real.

 

Figure 1: Cauchy-Riemann.
\begin{figure}\mbox{<br />
            \epsfig{file=Cauchy-Riemann.eps,height=5.5cm} }\end{figure}

Suppose that

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

 

$\displaystyle f'(z_0)$ $\displaystyle = \underset{\Delta x \rightarrow 0}{\text{lim}} \frac {[u(x_0+\Delta x, y_0)+i v((x_0+\Delta x, y_0)]-[u(x_0, y_0)+i v((x_0, y_0)]}{\Delta x}$    
$\displaystyle \quad$ $\displaystyle =\underset{\Delta x \rightarrow 0}{\text{lim}} \frac {u(x_0+\Delt...<br />
            ...ightarrow 0}{\text{lim}} \frac {v((x_0+\Delta x, y_0) - v((x_0, y_0)}{\Delta x}$    
$\displaystyle \quad$ $\displaystyle = u_x(x_0,y_0)+i v_x(x_0,y_0)$    



 

 

Suppose that

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

 

$\displaystyle f'(z_0)$ $\displaystyle = \underset{\Delta y \rightarrow 0}{\text{lim}} \frac {[u(x_0, y_0 + \Delta y)+i v((x_0, y_0+ \Delta y)]-[u(x_0, y_0)+i v((x_0, y_0)]}{i \Delta y}$    
$\displaystyle \quad$ $\displaystyle = \underset{\Delta y \rightarrow 0}{\text{lim}} \frac {u(x_0, y_0...<br />
            ...tarrow 0}{\text{lim}} \frac {v((x_0, y_0+ \Delta y) - v((x_0, y_0)}{i \Delta y}$    
$\displaystyle \quad$ $\displaystyle = - i v_y(x_0,y_0) + v_y(x_0,y_0)$    



 

 

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

 

\begin{displaymath}<br />
\begin{cases}<br />
\frac{\partial u}{\partial x}= \frac {\partial...<br />
...l v}{\partial x} = - \frac {\partial u}{\partial y}<br />
\end{cases}\end{displaymath}

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

 

\begin{displaymath}<br />
\begin{cases}<br />
u_x=v_y \ u_y=-v_x<br />
\end{cases}\end{displaymath}

 

 

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:

 

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


 

It follows that:

 

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


 

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:

 

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


 

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{displaymath}\begin{cases}2x=0 \ 2y=0\end{cases}\end{displaymath} , 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<br />
\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:

 

 

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


 

We compute the first partial derivatives:

 

\begin{displaymath}\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}\end{displaymath}    


 

We solve Cauchy-Riemann equations:

 

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


 

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<br />
\theta)$ .

 

\begin{displaymath}<br />
\begin{cases}<br />
\frac {\partial u}{\partial r} = \frac {1}{r} ...<br />
...ac {1}{r}<br />
\cdot \frac {\partial u}{\partial \theta}<br />
\end{cases}\end{displaymath} 

courtesy:ndp.jct.ac.il/tutorials/complex/node19.html


Dear Guest,
Spend a minute to Register in a few simple steps, for complete access to the Social Learning Platform with Community Learning Features and Learning Resources.
If you are part of the Learning Community already, Login now!
Tags:


0
Your rating: None

Posted by



Thu, 05/28/2009 - 10:13

Share

Collaborate