Chapter
1.5 Complex Numbers
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College Algebra by Example Series |
I. Convert the following to the standard form of the
complex numbers a + bi
Problem Examples | Standard form: a + bi |
(1) 5 + 3i | 5 + 3i (Already in standard form) |
(2) 4 - 5i | 4 + (-5)i |
(3) | (note is best written as
) |
(4) -6i | 0 + (-6)i (when a=0, bi is called a pure imaginary number) |
(5) 12 | 12 + 0i |
II. Add the following complex number:
Principles: (a + bi) + (c + di) = (a + c) + (b + d)i
(a + bi) - (c + di) = (a - c) + (b - d)i
(a + bi) + (-a - di) = 0 (where -a - di is called the additive inverse)
Examples - Add and substract as indicated
(1) (-7 - 3i) + (-4 + 4i) = (-7 -4) + (-3 + 4)i = -11 + i
(2) (-2 - 3i) - (-1 - i) = (-2 -(-1)) + (-3 - (-1))i = -1 - 2i
(3)
III. Mutipication and Division of Complex Numbers
Principles:
(1)
(2)
Use multiplication rules:
(1) (a + bi)(c + di) = a(c + di) + bi(c + di)
(2) (a + bi)2 = a2 + 2abi + (bi)2
(3) (a - bi)2 = a2 - 2abi + (bi)2
(4) (a + bi)(a - bi) = a2 - (bi)2 = a2 + b2 (conjugate used to simplify the quotient of 2 complex numbers)
Examples: Express each products in standard form:
(1) (4i)(3-2i) = 4i(3) + 4i(-2i) =12i - 8i2 = 12i - 8(-1) = 12i +8 = 8 + 12i
(2) (3 + 2i)(4 + 6i) = 3(4 + 6i) + 2i(4 + 6i) =12 + 18i + 8i + 12i2 = 12 + 26i + 12(-1) = 0 + 26i
= 0 + 26i
(3) (4 + 5i)(2 - 9i) = 4(2 - 9i) + 5i(2 - 9i) = 8 - 36i + 10i - 45i2 = 8 - 26i - 45(-1) = 53 - 26i
(4) (3 + 4i)2 = (3 + 4i)(3 + 4i) = 32 + 2(3)(4)i + (4i)2 = 9 + 24i + 16i2 = 9 + 24i + 16(-1)
= -7 + 24i
(5) (-1 - 2i)2 = (-1 - 2i)( -1 - 2i) = (-1)2 - 2(-1)(2)i + (2i)2 = 1 + 4i + 4i2 = 1 + 4i + 4(-1)
= -3 + 4i
(6) Conjugate: (3 + 4i)(3 - 4i) = (-3)2 - (4i)2 = 9 - 16i2 = 9 - 16(-1) = 9 + 16 = 25
= 25 + 0i
Examples: Express each quotients in standard form: (hint: use the conjugate)
(1)
(2)
(3)
(4)
Absolute Value
Example |
Example. Find
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Example. Find
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Example. Find First write in a + bi form (use conjugate)
Remember that So
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Example. Find
Example. Find
Example. Find ,First write in a + bi form
So
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