# Complex analysis

Basic reference on complex analysis.

Notes taken while taking a course .

## Basics

Disk
$$B_r(z_0) = \{ z \in \Complex \mid |z-z_0| < r \}$$
Circle
$$K_r(z_0) = \{ z \in \Complex \mid |z-z_0| = r \}$$
Interior point
For $$E \subset \Complex$$, $$z_0$$ is an interior point of $$E$$ if there is $$r > 0$$ such that $$B_r(z_0) \subset E$$.
Boundary point

For $$E \subset \Complex$$, $$b$$ is a boundary point of $$E$$ if for all $$r>0$$, $$E \cap B_r(b) \neq \emptyset$$ and $$E^\complement \cap B_r(b) \neq \emptyset$$.

The set of all boundary points of $$E$$ is the boundary set of $$E$$, denoted $$\partial E$$.

Open set
A set in $$\Complex$$ is open if all of its points are interior points.
Closed set
A set $$E$$ in $$\Complex$$ is closed if $$\partial E \subset E$$.
Closure
The closure of $$E$$ is $$\overline{E} = E \cup \partial E$$.
Interior
The interior of $$E$$ is the set $$\overset{\circ}{E}$$ of all interior points.

## Limit

If for all $$\varepsilon > 0$$ there is $$\delta > 0$$ such that $$|f(z) - c| < \varepsilon$$ whenever $$|z-z_0| < \delta$$, then

$\lim_{z \rightarrow z_0} f(z) = c$

$$f$$ is continuous at $$z_0$$ if

$\lim_{z \rightarrow z_0} f(z) = f(z_0)$

## Derivative

$\frac{d}{dz} f(z_0) = \lim_{z \rightarrow z_0} f(z)$

$$f$$ is analytic in an open set $$U \subset \Complex$$ if $$f$$ is differentiable for every $$z \in U$$. A function which is analytic in $$\Complex$$ is an entire function.

## Cauchy-Riemann equations

If $$f(z) = u(x,y) + i v(x,y)$$ for $$z = x+iy$$ and real functions $$u$$ and $$v$$,

$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$ $\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$

Also,

$\frac{d f}{d z} = \frac{\partial f}{\partial x} = -i\frac{\partial f}{\partial y}$

$$f=u+iv$$ is analytic in $$D$$ if and only if for any $$z \in D$$, $$\frac{\partial u}{\partial x}$$ and $$\frac{\partial v}{\partial y}$$ exist, are continuous, and satisfy the Cauchy-Riemann equations.

## Fundamental theorem of algebra

If $$a_0$$, $$\ldots$$, $$a_n$$ are complex numbers with $$a_n \neq 0$$, then the polynomial

$p(z) = \sum_{k=0}^n a_k z^k$

has $$n$$ complex roots $$z_1$$, $$\ldots$$, $$z_n$$, where

$p(z) = a_n \prod_{k=1}^n (z - z_k)$