Complex numbers

The real numbers can solve first-degree equations $ax+b=0,$ $a,b\in \mathbb{R} \text{ with } a\neq0.$ They cannot, however, solve every second-degree equation with real coefficients; for example the equation This calls for widening the notion of number so that, in the new set of numbers, every second-degree equation with real coefficients has solutions. That extension leads to the notion of a complex number.

A complex number has the form $z=a+ib,a,b\in R$ , where $i\int 2$ ; $Re\left(z\right)=a$ ; $Im\left(z\right)=b$ .

Representation of numbers in the complex plane:

$1+2i; 3+i; -2-3i; -2+i$

Plane vectors with their origin at the origin of the axes $(0,0)$ can be represented by a pair of real numbers $(a,b)$ (the tip of the vector). The complex number $z=a+ib$ is represented in the plane by the same point as the vector $(a,b)$ .

Operations with complex numbers.

Let $z\int_{1}$ and $z\int_{2}$

$z\int_{1}$ ;

$\bigcap 1=\bigcap a$ ; $\bigcap z\int_{2} z\int_{1} =\bigcap a\int_{1}$ ;

Conjugate complex numbers

$\prod z=\prod a+ib=a-ib$ is called the conjugate of the complex number $z=a+ib$ .

We have: a) $a=\bigcap z+\prod z$ ; $ib=\bigcap z-\prod z$ ; b) $z\in R$ if and only if $\prod z=z$ ;

c) $z$ is purely imaginary if and only if $\prod z=-z$ ; d) $\prod \alpha z=\alpha \cdot \prod z,\forall \alpha \in R$

Let $z=a+ib$ and $z'=a'+ib'$ be two complex numbers. We have:

a) $\prod z+z'=\prod z+\prod z'$ ; b) $\prod z\cdot z'=\prod z\cdot \prod z'$ ; c) $\bigcap z=\bigcap z\cdot \prod z'$ , $z'\neq0$ .

The modulus of a complex number

Let

For every complex number $z\in C,z=a+ib$ we have:

1); 2); 3); 4).

If $z\int_{1}$ , then: 5); 6) equation ;

The triangle inequality .

Trigonometric form of a complex number

Let $z\in C\int *$ , $z=a+ib$ . There exist unique and $ϕ\in \left[0,2\pi\right]$ such that: $z=\rho \left(cosϕ+isinϕ\right)$ ; $cosϕ=\bigcap a$ , $sinϕ=\bigcap b$ .

Let $z\int_{1}$ and $z\int_{2}$

$z\int_{1}$ $\bigcap 1$

$\bigcup_{z\int_{1} n} =$

De Moivre's formula: .

Expanding the left-hand side using the binomial theorem gives the following relation:

Euler's formula:

so for $\theta=\pi \Rightarrow$ .

Applications of complex numbers in geometry

The equation of the circle with centre $M\int_{0}$ and radius $r>0$ is

Let the points $M\int_{1}$ and $M\int_{2}$ . We have .

Let the points , , . Then

Given the points , , , , the lines and are perpendicular if and only if is purely imaginary, that is .

The points , , , are concyclic if and only if .

Two triangles and , with vertices of affix and respectively, are similar if and only if: .

The complex number $\varepsilon$

The cube roots of unity

The solutions of the equation $x\int 3$ are the complex roots of the polynomial $f=x\int 3$ , $f\in C\left[x\right]$

$x\int 3$ $x-1=0;$

$\Rightarrow$ $x\int_{1}$

$x\int 2$ $\Rightarrow$ $\Delta=b\int 2$

equation $\Rightarrow$ $x\int_{2,3}$

We adopt the following notation:

Properties:

$\varepsilon \int 3$

$\varepsilon \int 2$

$1\int -1$ , $\varepsilon \int -1$ , $(\varepsilon \int 2$


$\varepsilon \int 0\varepsilon \int 1\varepsilon \int 2$	$\varepsilon \int 3\varepsilon \int 4\varepsilon \int 5$	$\varepsilon \int 3k\varepsilon \int 3k+1\varepsilon \int 3k+2$

Trigonometric form of the complex number $\varepsilon$

$x\int 3$ $\Rightarrow x\int 3$

$\bigcup_{1} 3=cos(\bigcap 2k\pi)+isin(\bigcap 2k\pi),$ $k=0,1,2$

for k=0 we have: $\bigcup_{1} 3=cos(\bigcap 2k\pi)+isin(\bigcap 2k\pi)=cos(\bigcap 2\cdot0\cdot \pi)+isin(\bigcap 2\cdot0\cdot \pi)=cos(0)+isin(0)=1+i\cdot0=1$ for k=1 we have: $\bigcup_{1} 3=cos(\bigcap 2k\pi)+isin(\bigcap 2k\pi)=cos(\bigcap 2\cdot1\cdot \pi)+isin(\bigcap 2\cdot1\cdot \pi)=cos(\bigcap 2\pi)+isin(\bigcap 2\pi)=-\bigcap 1+i\bigcap \bigcup 32$

for k=2 we have: $\bigcup_{1} 3=cos(\bigcap 2k\pi)+isin(\bigcap 2k\pi)=cos(\bigcap 2\cdot2\cdot \pi)+isin(\bigcap 2\cdot2\cdot \pi)=cos(\bigcap 4\pi)+isin(\bigcap 4\pi)=-\bigcap 1-i\bigcap \bigcup 32$