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equality of complex numbers∗
pahio†
2013-03-22 0:16:28
The equality relation “=” among the complex numbers is determined as
consequence of the definition of the complex numbers as elements of the quotient
ring 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 i2 = −1.
⇐⇒
a1 + ib1 = a2 + ib2
a1 = a2 ∧ b1 = b2
(1)
This condition may as well be derived by using the field properties of C and the
properties of the real numbers:
a1 + ib1 = a2 + ib2 =⇒
a2 − a1 = −i(b2 − b1 )
=⇒ (a2 − a1 )2 = −(b2 − b1 )2
=⇒ (a2 − a1 )2 + (b2 − b1 )2 = 0
=⇒
a2 − a1 = 0, b2 − b1 = 0
=⇒
a1 = a2 ,
b1 = b2
The implication chain in the reverse direction is apparent.
If a + ib 6= 0, then at least one of the real numbers a and b differs from 0.
We can set
a = r cos ϕ,
b = r sin ϕ,
(2)
where r is a uniquely determined positive number and ϕ is an angle which is
uniquely determined up to an integer multiple of 2π. In fact, the equations (2)
yield
a2 + b2 = r2 (cos2 ϕ + sin2 ϕ) = r2 ,
whence
r=
p
a 2 + b2 .
∗ hEqualityOfComplexNumbersi
(3)
created: h2013-03-2i by: hpahioi version: h40210i Privacy
setting: h1i hTopici h30-00i
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1
Thus (2) implies
cos ϕ = √
a
,
+ b2
sin ϕ = √
a2
b
.
+ b2
a2
(4)
The equations (4) are compatible, since the sum of the squares of their right
sides is 1. So these equations determine the angle ϕ up to a multiple of 2π. We
can write the
Theorem. Every complex number may be represented in the polar form
r(cos ϕ + i sin ϕ),
where r is the modulus and ϕ 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 2π.
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