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
$\displaystyle a=r\cos\varphi,\qquad b=r\sin\varphi,$ | (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
$\displaystyle\cos\varphi=\frac{a}{\sqrt{a^{2}+b^{2}}},\qquad\sin\varphi=\frac{% b}{\sqrt{a^{2}+b^{2}}}.$ | (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 $2\pi$.