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

Conjugate

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

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

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