Derivation.

Definition 3.2.1   Let $ f$ be a function defined on a domain $ R$ in $ \mathbb{C}$ and let $ z_0$ be an interior point of $ R$ . The function $ f$ is derivable (or differentiable) at $ z_0$ if there exists a complex number $ \alpha$ such that

$\displaystyle \underset{z \rightarrow z_0}{\text{lim}} \frac {f(z)-f(z_0)}{z-z_0} = \alpha.$    

We denote this by $ f'(z_0)=\alpha$ .

Remark 3.2.2   Another formulation of Def. 2.1 is as follows:

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

Example 3.2.3   Let $ f(z)=z^n$ , where $ n \in \mathbb{N}$ . Then, for every $ z_0 \in \mathbb{C}$ :

$\displaystyle \frac {f(z)-f(z_0)}{z-z_0}=\frac {z^n-z_0^n}{z-z_0}=z^{n-1}+z^{n-2}z_0+ z^{n-3}z_0^2+ \dots + z_0^{n-1}$    

from what follows:

$\displaystyle \underset{z \rightarrow z_o}{\lim}\frac {f(z)-f(z_0)}{z-z_0} = \u...
...rrow z_o}{\lim} (z^{n-1}+z^{n-2}z_0+ z^{n-3}z_0^2+ \dots + z_0^{n-1})=nz^{n-1}.$    

The function $ f$ is differentiable at every point $ z_0$ and $ f'(z_0)=nz_0^{n-1}$ .

The algebra of derivable functions of a complex variable is (formally) identical to the algebra of derivable functions of a real variable seen in Calculus.

Proposition 3.2.4   Let $ f$ and $ g$ be two functions defined on a neighborhood of $ z_0$ . We suppose that $ f$ and $ g$ are continuous at $ z_0$ .
(i)
$ f+g$ is differentiable at $ z_0$ and $ (f+g)'(z_0) = f'(z_0)+g'(z_0)$ .
(ii)
$ fg$ is differentiable at $ z_0$ and we have:

$\displaystyle (fg)'(z_0)=f'(z_0)g(z_0) + f(z_0)g'(z_0).$    

(iii)
If $ g(z_0) \neq 0$ , then $ 1/g$ is derivable at $ z_0$ and we have $ (1/g)'(z_0)= - g'(z_0)/(g(z_0)^2$ .
(iv)
If $ g(z_0) \neq 0$ ,then $ f/g$ is derivable at $ z_0$ and we have:

$\displaystyle \left( \frac {f}{g} \right)'(z_0)= \frac {f'(z_0)g(z_0)-f(z_0)g'(z_0)}{g(z_0)^2}$    

Example 3.2.5   If $ f(z)=1/z$ , then $ f$ is differentiable at every point of $ \mathbb{C}^*$ and $ f'(z)=-1/z^2$ .

Proposition 3.2.6   If $ f$ is derivable at $ z_0$ and if $ g$ is derivable at $ f(z_0)$ , then $ gof$ is derivable at $ z_0$ and we have:

$\displaystyle (gof)'(z_0)=g'(f(z_0)) \cdot f'(z_0).$    

As easy consequences of Proposition 2.4, we have Corollary 2.7 and Corollary 2.8.

Corollary 3.2.7        A polynomial function is differentiable on the whole of $ \mathbb{C}$ .

Corollary 3.2.8        A rational function is differentiable at every point of its domain of definition

Example 3.2.9   The function $ f$ such that $ f(z)= ((1+i)z-2i)/(z^2+1)$ is differentiable on $ \mathbb{C} - \{ -i, i \}$ and we have:

$\displaystyle \forall z \in \mathbb{C} - \{ -i, i \}, \; f'(z)= \frac {(1+i)(z^2+1)-((1+i)z-2i)2z}{(z^2+1)^2} = \frac {-(1+i)z^2+4iz+1+i}{(z^2+1)^2}.$    

Actually, we do not dispose of enough functions in order to display more interseting examples. For that purpose, we shall wait until next chapter (i.e. Chapter chapter analytic functions).

Noah Dana-Picard 2007-12-24