Download Math 0031 Section 3.1 Page 1 of 4 3.1 Complex Numbers The

Math 0031
Section 3.1
Page 1 of 4
3.1 Complex Numbers
The Number i
The number i is defined such that i2 = −1.
√
In this course we also assume that i = −1.
Example Express each number in terms of i.
(a)
√
−36.
Solution:
(b)
√
p
√
√
√
−36 = (−1) · 36 = −1 · 36 = i · 6 = 6i
−32.
Solution:
p
√ √
√
√
√
√
√
−32 = (−1) · 2 · 16 = −1 · 2 · 16 = i · 2 · 4 = 4i 2 = 4 2 i
Complex Numbers
A complex number is a number of the form a + bi, where a and b are real numbers.
The number a is said to be the real part of a+bi and the number b is said to be the imaginary
part of a + bi.
If b = 0 then a + bi = a + 0i = a is a real number. So every real number is a complex number.
If a = 0 then a + bi = 0 + bi = bi is an imaginary number (or a pure imaginary number).
Note: b is a real number and bi is an imaginary number. Sometimes i is also called an
imaginary number because i = 1i.
Addition and Subtraction
(a + bi) + (c + di) = (a + c) + (b + d)i.
(a + bi) − (c + di) = (a − c) + (b − d)i.
Example Add or subtract each of the following.
(a)
(2 + 6i) + (7 − 11i)
Solution:
(b)
(2 + 6i) + (7 − 11i) = (2 + 7) + (6 + (−11))i = 9 − 5i
(2 + 6i) − (7 − 11i)
Solution:
(2 + 6i) − (7 − 11i) = (2 − 7) + (6 − (−11))i = −5 + 17i
Math 0031
Section 3.1
Page 2 of 4
Multiplication
(a + bi) · (c + di) = (ac − bd) + (ad + bc)i.
Indeed, (a + bi) · (c + di) = ac + adi + bci + bdi2 = ac + (ad + bc)i − bd = (ac − bd) + (ad + bc)i.
√
√
Note: In real numbers
1√= 1. In complex numbers
1 can be both 1 and −1 because
2
2
(−1) = 1 and 1 = 1. So 1 = ±1.
Due to this duality we have, for example
√ √
√
√
√
√
−4 · −9 = −1 · 4 · −1 · 9 = i · 2 · i · 3 = 6i2 = −6
p
√
√
√
and −4 · −9 = (−4)(−9) = 36 = 6
The ambiguity comes from the fact that
p
√ √
√
√
√
√
√
√
−4 · −9 = −1 · 4 · −1 · 9 = (−1)(−1) · 36 = 1 · 6 = ±1 · 6 = ±6.
For the sake of certainty in this course we define that for positive reals a and b the only correct
multiplication of two imaginary numbers in the radical notation is
√
√
√
√
√
√
√
√ √
√
−a · −b = −1 · a · −1 · b = i · a · i · b = i2 ab = − ab
Example Multiply and simplify each of the following.
(a)
√
−10 ·
√
−8
Solution:
(b)
(2 + i)(7 − 3i)
Solution:
(c)
√
√
√
√
√
√
√
√
√
−10 · −8 = −1 · 10 · −1 · 8 = i2 10 · 8 = − 5 · 16 = −4 5
(2 + i)(7 − 3i) = (14 − (−3)) + (−6 + 7)i = 17 + i
(2 + 3i)2
Solution:
(2 + 3i)2 = (2 + 3i)(2 + 3i) = (4 − 9) + (6 + 6)i = −5 + 12i
Powers of i
It is easy to calculate the following powers of i
i2 = −1.
i3 = i2 · i = −1i = −i.
i4 = i2 · i2 = (−1)(−1) = 1.
i5 = i4 · i = 1 · i = i.
and the cycle is closed.
Math 0031
Section 3.1
Page 3 of 4
Therefore for positive integers n, m, and k such that k < 4 and n = 4m + k we have
in = i4m+k = i4m · ik = (i4 )m · ik = 1m · ik = ik
Example Simplify each of the following.
(a)
i49
Solution:
(b)
49 = 4 · 12 + 1. Therefore i49 = i1 = 1
i91
Solution:
91 = 4 · 22 + 3. Therefore i91 = i3 = −i
Conjugates and Division
Conjugate of a Complex Number
The conjugate of a complex number a + bi is a − bi. The numbers a + bi and a − bi are
complex conjugates.
Example Find the conjugate to the number −12 − 3i.
Solution:
It is −12 − (−3)i = 12 + 3i
Example Find the product of the number 2 − 3i and its conjugate.
Solution:
The conjugate is 2 + 3i. Then (2 − 3i)(2 + 3i) = (4 − (−9)) + (6 − 6)i = 13
Product of conjugates
(a + bi)(a − bi) = a2 + b2
Division
(a + bi) ÷ (c + di) =
ac + bd bc − ad
+ 2
i.
c2 + d 2
c + d2
Indeed,
(a + bi) ÷ (c + di) =
a + bi
(a + bi)(c − di)
(ac + bd) + (bc − ad)i
ac + bd bc − ad
=
=
= 2
+ 2
i.
2
2
c + di
(c + di)(c − di)
c +d
c + d2
c + d2
Example Divide and simplify each of the following.
(a)
(2 + i) ÷ (3 + 5i)
Math 0031
Solution:
(b)
Section 3.1
(2 + i) ÷ (3 + 5i) =
6+5
3 − 10
+
i=
32 + 52 32 + 52
Page 4 of 4
11
34
−
7
34
i
(−2 + 5i) ÷ (7 − 4i)
Solution:
(−2 + 5i) ÷ (7 − 4i) =
−14 − 20 35 − (−2)(−4)
+
+
i = − 34
65
2
2
2
2
7 +4
7 +4
27
65
i