First, let's us recall some notion in real analysis. Given a function $f: \R \to \R$, we can talk about its derivative at a point $x_0 \in \R$. It is defined as $$ f'(x_0) = \lim_{\epsilon \to 0} \frac{f(x_0 + \epsilon) - f(x_0)} {\epsilon}. $$ If the derivative $f'(x_0)$ exists, then we say the function $f$ is differentiable at $x_0$. For example, $f(x) =|x|$ is not differentiable at $x=0$. If the derivative $f'(x_0)$ exists for all $x_0 \in \R$, then we say $f$ is differentiable on $\R$.

Ex 1 $$ f(x) = \begin{cases} x & x \geq 0 \cr 2 x & x \leq 0 \end{cases} $$ Is it differentiable at $x=0$? Can you plot $f'(x)$?

Ex 2 $$ f(x) = \begin{cases} x^2 & x \geq 0 \cr 2 x^2 & x \leq 0 \end{cases} $$ Is it differentiable at $x=0$? Can you plot $f'(x)$? (this is an example, where the function $f(x)$ is differentiable, but the derivative $f'(x)$ is not continuous, hence we cannot define $f(x)$ at $x=0$, hence $f(x)$ is not a smooth function on $\R$). ==== Complex function Differentiability ==== On the face value, the definition for complex differentiability is just replacing the word 'real' by 'complex', in the above section. Definition: we say $f: \C \to \C$ is differentiable at $z_0 \in \C$, if $$ f'(z_0) = \lim_{\epsilon \to 0} \frac{f(z_0 + \epsilon) - f(z_0)} {\epsilon} \quad \text{ exists.} $$ Remark: Here $\epsilon$ can go to $0$ in different directions, we can let $\epsilon = r e^{i\theta}$ for fixed $\theta$ and positive $r \to 0$. The condition that derivative exists really means no matter how $\epsilon$ approach $0$, the above limit exists. Let's see some example. * $ f(z) = z^3 $ at $z_0$, what is $f'(z_0)$? * $ f(z) = |z|^2 $ at $z_0 = 0$, at $|z_0|=1$. Does $f'(z_0)$ exist? Definition: we say a subset $\Omega \In \C$ is an 'open subset of $\C$', or simply 'open', if for any point $p \in \Omega$, we can enlarge it to a small ball $B_\epsilon(p) \In \Omega$, where $B_\epsilon(p) = \{ z \mid |z-p| < \epsilon \}$. Example: the strip $| Re(z) | < 1$ is open; the line $Re z = 1$ is not open. Definition: We say a function $f: \Omega \to \C$ is an analytic function (aka holomorphic) on $\Omega$ if for any $p \in \Omega$, $f'(p)$ exists. Example: * $f(z) = 1/z$ is holomorphic on $\Omega = \C \RM \{0\}$. * $f(z) = 1/ (z^2+1)$ is holomorphic on $\Omega = \{z \in \C \mid z^2 + 1 \neq 0 \} = \C \RM \{ \pm i \}$. ===== Taylor expansion ===== Suppose $f: \Omega \to \C$ is a holomorphic function on $\Omega$. Then the derivative $f'(p)$ exists for every point $p$ in $\Omega$ by definition. Furthermore, $f': \Omega \to \C$ itself is a holomorphic function (a non-trivial result, not true for real differentiable function). Hence, we can differentiate $f$ anytimes we want on $\Omega$. The Taylor expansion of $f$ centered at point $p \in \Omega$ is the following identity. If $B_r(p) \In \Omega$, then for any $z \in B_r(p)$, we have $$ f(z) = f(p) + f'(p) (z-p) + f(p) \frac{ (z-p)^2}{2!} + \cdots + f^{(n)}(p) \frac{ (z-p)^2}{n!} + \cdots. $$