User:Aleksander Stos/ComplexNumberAdvanced

This is an experimental draft. For a brief description of the project and motivations click here.

Complex numbers are defined as ordered pairs of reals:

$\mathbb {C} =\{(a,b)\colon a,b\in \mathbb {R} \}.$

Such pairs can be added and multiplied as follows

• addition: $(a,b)+(c,d)=(a+c,b+d)$
• multiplication: $(a,b)(c,d)=(ac-bd,bc+ad)$

$\scriptstyle \mathbb {C}$ with the addition and the multiplication is the field of complex numbers.

To perform basic computations it is convenient to introduce the imaginary unit, i=(0,1). It has the property $\scriptstyle i^{2}=-1.$ Any complex number $z=(a,b)$ can be written as $z=a+bi$ (this is often called the algebraic form) and vice-versa. This makes the multiplication natural:

$(a+bi)(c+di)=(ac-bd)+(bc+ad)i.$

Any complex number $\scriptstyle z=a+bi\not =0$ has its inverse. Actually,

${\frac {1}{z}}={\frac {a}{a^{2}+b^{2}}}-i{\frac {b}{a^{2}+b^{2}}}.$

The square root of number in the denominator in the above formula is called the modulus of z and denoted by $|z|$,

$|z|={\sqrt {a^{2}+b^{2}}}.$

We have for any two complex numbers $z_{1}$ and $z_{2}$

• $|z_{1}\cdot z_{2}|=|z_{1}|\cdot |z_{2}|;$
• $|{\frac {z_{1}}{z_{2}}}|={\frac {|z_{1}|}{|z_{2}|}},$ provided $z_{2}\not =0.$
• ${\big |}|z_{1}|-|z_{2}|{\big |}\leq |z_{1}+z_{2}|\leq |z_{1}|+|z_{2}|$

For $z=a+bi$ we define also ${\bar {z}}$, the conjugate, by ${\bar {z}}=a-bi.$ Then we have

• ${\bar {z}}_{1}\pm {\bar {z}}_{2}={\overline {(z_{1}\pm z_{2})}};$
• ${\bar {z}}_{1}\cdot {\bar {z}}_{2}={\overline {(z_{1}\cdot z_{2})}};$
• ${\overline {\left({\frac {z_{1}}{z_{2}}}\right)}}={\frac {{\bar {z}}_{1}}{{\bar {z}}_{2}}},$ provided $z_{2}\not =0;$
• $z{\bar {z}}=|z|^{2}.$