tl;dr

A function $f: \C \to \C \cup \{\infty\}$ is holomorphic at $z_0$ if there exists $\epsilon>0$, and complex numbers $a_0,a_1,\cdots$ that for all $|z - z_0|<\epsilon$, we can write $f(z) = \sum_{n=0}^\infty a_n (z-z_0)^n.$

A function $f: \C \to \C \cup \{\infty\}$ is meromorphic at $z_0$ with order $m \geq 1$ pole, if there exists $\epsilon>0$, and complex numbers $a_{-m}, \cdots , a_0,a_1,\cdots$ that for all $0 < |z - z_0|<\epsilon$, we can write $f(z) = \sum_{n=-m}^\infty a_n (z-z_0)^n.$

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 $\R$).

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 $\Omega = \{z \in \C \mid z^2 + 1 \neq 0 \} = \C \RM \{ \pm i \}$.

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 $f(z) = f(p) + f'(p) (z-p) + f”(p) \frac{ (z-p)^2}{2!} + \cdots + f^{(n)}(p) \frac{ (z-p)^2}{n!} + \cdots.$

Suppose $f: \C \RM \{0\} \to \C$ is holomorphic function, you cannot help but wonder, what goes wrong at $0$? You might find

• a removable singularity (i.e. a false alarm, $f$ is all good at $z=0$)
• a pole, $z^n f(z)$ is holomorphic
• an essential singularity: anything else, like $e^{1/z}$, $e^{1/z^2}$.

If you had a pole, we can apply that Taylor expansion formula (at $z=0$) to $z^n f(z)$, then divide out by $z$.

Find the first two terms in these expansions.

1. Taylor expand $(z+1)(z+2)$ around $z=3$.

2. Laurent expand $1/[(z-1)(z-2)]$ around $z=1$. And do it again, this time around $z=2$.

2.5 (Optional) Laurent expand $e^{1/z + z}$ around $z=0$.

3. You may have heard about Cauchy Riemann equation: let $f(z)$ be a $\C$-valued function on $\C$, and let $z = x+iy$, $f = u+iv$, then we can view $u,v$ as real valued functions depending on $x,y$.

If $f$ is holomorphic, then we have $\frac{\d u(x,y)}{\d x} = \frac{\d v(x,y)}{\d y}, \quad \frac{\d v(x,y)}{\d x} = - \frac{\d u(x,y)}{\d y}$

Your task: either prove this in general if you feel strong, or verify that this is true for your favorite holomorphic function (don't choose $f$ to be a constant, too boring)