Analytic functions.

Definition 3.5.1   A function $f$ is analytic at $z_0$ if the following conditions are fulfilled:

1. $f$ is derivable at $z_0$ ;
2. There exists a neighborhood $V$ of $z_0$ such that $f$ is derivable at every point of $V$ .

The simplest open neighborhood of a point $z_0$ is an open ball centered at $z_0$ ; we will generally use such balls.

Definition 3.5.2   Let $f$ be a function defined over an open subset $V$ of $\mathbb{C}$ . the function $f$ is analytic over $V$ if it is analytic at each point of $V$ .

A function which is analytic over the whole of $\mathbb{C}$ is called an entire function.

Proposition 3.5.3       

1. A polynomial function is analytic at every point of $\mathbb{C}$ , i.e. is an entire function.
2. A rational function is analytic at each point of its domain.

Proof.     

1. A polynomial function is an entire function, as a consequence of Proposition 2.4 and Ex. 2.3.
2. We use once again Proposition 2.4, together with the previous alinea, as a rational function is the quotient of two polynomial functions.

$\qedsymbol$

Example 3.5.4   Let $f(z)=\vert z\vert^2$ , i.e. $f(z)=x^2+y^2$ (i.e. $u(x,y)=x^2+y^2$ and $v(x,y)=0$ ). Then:

$$u_x=2x, \; u_y=2y, \; v_x=v_y=0.$$

C-R equations mean here $$\begin{cases}2x=0 \ 2y=0\end{cases}$$ , i.e. $(x,y)=(0,0)$ . Thus, the origin is the only point where $f$ can be derivable and $f$ is nowhere analytic.

Example 3.5.5   We use the function and the results of Example 3.5. For any point $z_0$ on one of these axes, any open ball centered at $z_0$ contains points where $f$ is not differentiable. Thus, $f$ is not analytic at any point.

Example 3.5.6   If $f(z)=\frac {1}{z^2+1}$ , the function $f$ is analytic at every point of $\mathbb{C} - \{ -i, i \}$ :

• With the notations of 3.1, we have:

  $$\begin{cases}u(x,y)= \frac {x^2-y^2+1}{(x^2+y^2+1)^2}\ v(x,y)= \frac {-2xy}{(x^2+y^2+1)^2} \end{cases}$$

  
  
  Thus:

  $$\begin{cases}u_x=v_y=\frac {-2x(x^2-y^2+1)^2(3y^2+1)}{x^4+2x^... ...(x^2+1)-3x^4-2x^2+1}{y^4+2y^2(x^2-1)+x^4+2x^2+1)^2} \end{cases}$$

  
  

• Cauchy-Riemann equations are verified by the first partial derivatives, and these derivatives are continuous functions on their domain. By Thm 3.4, the function $f$ is differentiable at every point of its domain $\mathbb{C}-{i,-i}$ .
• As $f$ is not defined at $i$ and at $-i$ , $f$ is not entire.

Example 3.5.7   Let $f(z)=x^2+iy^2$ , i.e. $U(x,y)=x$ and $v(x,y)=y$ . We have:

$$\begin{cases}u_x=2x \ u_y=0 \end{cases} \qquad \text{and} \qquad \begin{cases}v_x=0 \ v_y=2y \end{cases}$$

Thus, Cauchy-Rieman equations are verified if, and only if, $2x=2y$ , i.e. $x=y$ . It follows that $f$ can be derivable only on the line whose equation is $x-y=0$ . For any point $z_0$ on this line, every non empty open ball centered at $z_0$ has points out of the line, therefore $f$ cannot be analytic anywhere (see Figure 3).

Figure 3: Why a function is nowhere analytic.