PlanetMath (more info)
 Math for the people, by the people. Sponsor PlanetMath
Encyclopedia | Requests | Forums | Docs | Wiki | Random | RSS  
Login
create new user
name:
pass:
forget your password?
Main Menu
sections
Encyclopædia
Papers
Books
Expositions

meta
Requests (236)
Orphanage
Unclass'd (1)
Unproven (540)
Corrections (46)
Classification

talkback
Polls
Forums
Feedback
Bug Reports

downloads
Snapshots
PM Book

information
News
Docs
Wiki
ChangeLog
TODO List
Copyright
About
Owner confidence rating: Very high Entry average rating: Very high
[parent] equality of complex numbers (Topic)

The equality relation ``='' among the complex numbers is determined as consequence of the definition of the complex numbers as elements of the quotient ring $\mathbb{R}/(X^2\!+\!1)$ , which enables the interpretation of the complex numbers as the ordered pairs $(a,\,b)$ of real numbers and also as the sums $a\!+\!ib$ where $i^2 = -1$ .

$\displaystyle a_1+ib_1 = a_2+ib_2 \quad \Longleftrightarrow \quad a_1 = a_2\; \wedge\; b_1 = b_2$ (1)

This condition may as well be derived by using the field properties of $\mathbb{C}$ and the properties of the real numbers:
$\displaystyle a_1+ib_1 = a_2+ib_2\;$ $\displaystyle \implies\; \;a_2-a_1 = -i(b_2-b_1)$    
  $\displaystyle \implies\; (a_2-a_1)^2 = -(b_2-b_1)^2$    
  $\displaystyle \implies\; (a_2-a_1)^2+(b_2-b_1)^2 = 0$    
  $\displaystyle \implies\; \;a_2-a_1 = 0, \;\; b_2-b_1= 0$    
  $\displaystyle \implies\; \;a_1 = a_2, \;\;\; b_1 = b_2$    

The implication chain in the reverse direction is apparent.

If $a+ib \neq 0$ , then at least one of the real numbers $a$ and $b$ differs from 0. We can set

$\displaystyle a = r\cos\varphi, \quad b = r\sin\varphi,$ (2)

where $r$ is a uniquely determined positive number and $\varphi$ is an angle which is uniquely determined up to an integer multiple of $2\pi$ . In fact, the equations (2) yield $$a^2+b^2 = r^2(\cos^2\varphi+\sin^2\varphi) = r^2,$$ whence
$\displaystyle r = \sqrt{a^2+b^2}.$ (3)

Thus (2) implies
$\displaystyle \cos\varphi = \frac{a}{\sqrt{a^2+b^2}}, \quad \sin\varphi = \frac{b}{\sqrt{a^2+b^2}}.$ (4)

The equations (4) are compatible, since the sum of the squares of their right sides is 1. So these equations determine the angle $\varphi$ up to a multiple of $2\pi$ . We can write the

Theorem. Every complex number may be represented in the polar form $$r(\cos\varphi+i\sin\varphi),$$ where $r$ is the modulus and $\varphi$ the argument of the number. Two complex numbers are equal if and only if they have equal moduli and, if the numbers do not vanish, their arguments differ by a multiple of $2\pi$ .



"equality of complex numbers" is owned by pahio.
(view preamble | get metadata)

View style:

See Also: modulus of a complex number, argument of product and quotient, complex logarithm


This object's parent.
Log in to rate this entry.
(view current ratings)

Cross-references: vanish, argument, modulus, polar form, theorem, squares, implies, equations, multiple, integer, angle, number, positive, implication, properties, field, sums, real numbers, ordered pairs, quotient ring, complex numbers, consequence, equality relation
There are 3 references to this entry.

This is version 3 of equality of complex numbers, born on 2008-01-24, modified 2008-01-24.
Object id is 10210, canonical name is EqualityOfComplexNumbers.
Accessed 1597 times total.

Classification:
AMS MSC: 30-00 (Functions of a complex variable :: General reference works )
Pending Errata and Addenda
None.
Discussion
Style: Expand: Order:
forum policy

No messages.

Interact
post | correct | update request | add example | add (any)