3.1.10 Complex Numbers
The following is a review of Algebra-I concerning complex numbers. Recall that i = − 1 i = \sqrt{-1} i = − 1 a + b i a + bi a + bi
Complex Conjugate: a − b i a - bi a − bi Complex Modulus: a 2 + b 2 \sqrt{a^2 + b^2} a 2 + b 2 Complex Argument: arctan b a \arctan \frac{b}{a} arctan a b
The only questions that are usually asked on the number sense test is multiplying two complex numbers and rationalizing a complex number. Let’s look at examples of both:
Multiplication: ( a + b i ) ⋅ ( c + d i ) = ( a c − b d ) + ( a d + b c ) i (a + bi) \cdot (c + di) = (ac - bd) + (ad + bc)i ( a + bi ) ⋅ ( c + d i ) = ( a c − b d ) + ( a d + b c ) i
Example: ( 3 − 2 i ) ⋅ ( 4 + i ) = a + b i , a + b = (3 - 2i) \cdot (4 + i) = a + bi, a + b = ( 3 − 2 i ) ⋅ ( 4 + i ) = a + bi , a + b = Solution: a = 3 ⋅ 4 + 2 ⋅ 1 = 14 a = 3 \cdot 4 + 2 \cdot 1 = 14 a = 3 ⋅ 4 + 2 ⋅ 1 = 14 b = 3 ⋅ 1 + ( − 2 ) ⋅ 4 = − 5 b = 3 \cdot 1 + (-2) \cdot 4 = -5 b = 3 ⋅ 1 + ( − 2 ) ⋅ 4 = − 5 a + b = 14 − 5 = 9 a + b = 14 - 5 = 9 a + b = 14 − 5 = 9
Rationalizing: ( a + b i ) − 1 = a − b i a 2 + b 2 (a + bi)^{-1} = \frac{a - bi}{a^2 + b^2} ( a + bi ) − 1 = a 2 + b 2 a − bi
Example: ( 3 − 4 i ) − 1 = a + b i , a − b = ? (3 - 4i)^{-1} = a + bi, a - b = ? ( 3 − 4 i ) − 1 = a + bi , a − b = ? Solution: ( 3 − 4 i ) − 1 = 3 + 4 i 3 2 + 4 2 ⇒ a = 3 25 (3 - 4i)^{-1} = \frac{3 + 4i}{3^2 + 4^2} \Rightarrow a = \frac{3}{25} ( 3 − 4 i ) − 1 = 3 2 + 4 2 3 + 4 i ⇒ a = 25 3 b = 4 25 b = \frac{4}{25} b = 25 4 a − b = 3 25 − 4 25 = − 1 25 a - b = \frac{3}{25} - \frac{4}{25} = -\frac{1}{25} a − b = 25 3 − 25 4 = − 25 1
The following are some more practice problems about Complex Numbers:
Problem Set 3.1.10
( 4 − i ) 2 = a + b i , a = (4 - i)^2 = a + bi, a = ( 4 − i ) 2 = a + bi , a =
( 6 − 5 i ) ( 6 + 5 i ) = (6 - 5i)(6 + 5i) = ( 6 − 5 i ) ( 6 + 5 i ) =
The conjugate of ( 4 i − 6 ) (4i - 6) ( 4 i − 6 ) is a + b i , a = a + bi, a = a + bi , a =
( 5 + i ) 2 = a + b i , a = (5 + i)^2 = a + bi, a = ( 5 + i ) 2 = a + bi , a =
( 9 − 3 i ) ( 3 + 9 i ) = a + b i , a = (9 - 3i)(3 + 9i) = a + bi, a = ( 9 − 3 i ) ( 3 + 9 i ) = a + bi , a =
( 8 + 3 i ) ( 3 − 8 i ) = a + b i , a = (8 + 3i)(3 - 8i) = a + bi, a = ( 8 + 3 i ) ( 3 − 8 i ) = a + bi , a =
( 2 + 3 i ) ÷ ( 2 i ) = a + b i , a = (2 + 3i) \div (2i) = a + bi, a = ( 2 + 3 i ) ÷ ( 2 i ) = a + bi , a =
( 3 − 4 i ) ( 3 + 4 i ) = (3 - 4i)(3 + 4i) = ( 3 − 4 i ) ( 3 + 4 i ) =
( 24 − 32 i ) ( 24 + 32 i ) = (24 - 32i)(24 + 32i) = ( 24 − 32 i ) ( 24 + 32 i ) =
( 5 + 12 i ) 2 = a + b i , a + b = (5 + 12i)^2 = a + bi, a + b = ( 5 + 12 i ) 2 = a + bi , a + b =
( 3 − 5 i ) ( 2 − 5 i ) = a + b i , a + b = (3 - 5i)(2 - 5i) = a + bi, a + b = ( 3 − 5 i ) ( 2 − 5 i ) = a + bi , a + b =
( 2 − 5 i ) ( 3 + 5 i ) = a + b i , a = (2 - 5i)(3 + 5i) = a + bi, a = ( 2 − 5 i ) ( 3 + 5 i ) = a + bi , a =
( 2 − 5 i ) ( 3 − 4 i ) = a + b i , a − b = (2 - 5i)(3 - 4i) = a + bi, a - b = ( 2 − 5 i ) ( 3 − 4 i ) = a + bi , a − b =
( 4 − 3 i ) ( 2 − i ) = a + b i , a − b = (4 - 3i)(2 - i) = a + bi, a - b = ( 4 − 3 i ) ( 2 − i ) = a + bi , a − b =
( 2 + 7 i ) ( 2 − 7 i ) = a + b i , a − b = (2 + 7i)(2 - 7i) = a + bi, a - b = ( 2 + 7 i ) ( 2 − 7 i ) = a + bi , a − b =
( 2 + 3 i ) ( 4 + 5 i ) = a + b i , a = (2 + 3i)(4 + 5i) = a + bi, a = ( 2 + 3 i ) ( 4 + 5 i ) = a + bi , a =
( 3 + 4 i ) 2 = a + b i , a = (3 + 4i)^2 = a + bi, a = ( 3 + 4 i ) 2 = a + bi , a =
The modulus of 14 + 48 i 14 + 48i 14 + 48 i is
( 2 − 5 i ) 2 = a + b i , a + b = (2 - 5i)^2 = a + bi, a + b = ( 2 − 5 i ) 2 = a + bi , a + b =
( 5 + 4 i ) ( 3 + 2 i ) = a + b i , a = (5 + 4i)(3 + 2i) = a + bi, a = ( 5 + 4 i ) ( 3 + 2 i ) = a + bi , a =
( 0 + 4 i ) 2 = a + b i , b = (0 + 4i)^2 = a + bi, b = ( 0 + 4 i ) 2 = a + bi , b =
( 4 + 5 i ) ( 4 − 5 i ) = (4 + 5i)(4 - 5i) = ( 4 + 5 i ) ( 4 − 5 i ) =
The modulus of ( 11 + 60 i ) 2 (11 + 60i)^2 ( 11 + 60 i ) 2 is
( 0 − 3 i ) 5 = a + b i , b = (0 - 3i)^5 = a + bi, b = ( 0 − 3 i ) 5 = a + bi , b =
( 3 − 5 i ) ( 2 + i ) = a + b i , a + b = (3 - 5i)(2 + i) = a + bi, a + b = ( 3 − 5 i ) ( 2 + i ) = a + bi , a + b =
( 4 − 2 i ) ( 3 − i ) = a + b i , a + b = (4 - 2i)(3 - i) = a + bi, a + b = ( 4 − 2 i ) ( 3 − i ) = a + bi , a + b =
( 1 + i ) 9 = (1 + i)^9 = ( 1 + i ) 9 =
( 2 + 3 i ) ÷ ( 3 − 2 i ) = a + b i , b = (2 + 3i) \div (3 - 2i) = a + bi, b = ( 2 + 3 i ) ÷ ( 3 − 2 i ) = a + bi , b =
( 2 − 3 i ) ÷ ( 3 − 2 i ) = a + b i , a = (2 - 3i) \div (3 - 2i) = a + bi, a = ( 2 − 3 i ) ÷ ( 3 − 2 i ) = a + bi , a =
( 3 + 4 i ) ÷ ( 5 i ) = a + b i , a + b = (3 + 4i) \div (5i) = a + bi, a + b = ( 3 + 4 i ) ÷ ( 5 i ) = a + bi , a + b =
The modulus of ( 24 + 7 i ) 2 (24 + 7i)^2 ( 24 + 7 i ) 2 is
( 3 i − 2 ) ÷ ( 3 i + 2 ) = a + b i , b = (3i - 2) \div (3i + 2) = a + bi, b = ( 3 i − 2 ) ÷ ( 3 i + 2 ) = a + bi , b =
The modulus of ( 5 + 12 i ) 2 (5 + 12i)^2 ( 5 + 12 i ) 2 is