Complex numbers

Fundamental theorem of algebra

NOTE

a polynomial of degree $n\ge 1$ has exactly $n$ complex roots.

This implies that complex numbers form an algebraically closed field see proofs:

• using Complex Analysis: Fundamental Theorem of Algebra
• using Algebraic Topology: Fundamental Theorem of Algebra

Calculation

Conjugate

$\overline{(a+ib)}=a-ib$ $v^{\ast }={\bar{v}}^{\top }$ for $v\in {\mathbb{C}}^n$ $z^{-1}=\frac{\bar{z}}{|z{|}^2}=\frac{a-bi}{a^2+b^2}$ useful identity: $i^{-1}=-i$

squared absolute value

$|z{|}^2=z\cdot \bar{z}=a^2+b^2$

Example equation

$$z^n=1$$

$z_j=\cos \left(\frac{2\pi }{n}j\right)+i\sin \left(\frac{2\pi }{n}j\right)=e^{i\cdot 2\pi \cdot (j/n)}$

Vector

$v\in {\mathbb{C}}^n⟹v=\left(\begin{array}{c}{v}_1\\ v_2\\ ⋮\\ v_n\end{array}\right)$ with $v_i\in \mathbb{C}$

Scalar product

$⟨v,w⟩=w^{\ast }v=\overline{v^{\ast }w}$

Matrix

For $A\in {\mathbb{C}}^{m\times n}$

Hermitian conjugate

Let $A^{\ast }={\bar{A}}^{\top }$ with $\bar{A}=\left[\begin{array}{c}{\bar{A}}_{i,j}{\forall }_{i,j}\end{array}\right]\in {\mathbb{C}}^{m\times n}$

$(AB{)}^{\ast }=B^{\ast }{A}^{\ast }$

Isometry

Let $u$ be a unit complex number with $|u|=1$ distance from $uv$ to $uw$ is the same as the distance from $v$ to $w$ $|uv-uw|=|u(v-w)|=|u||v-w|=|v-w|$ (by multiplicative absolute value)

Isometry transformations preserves distance and angles