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Published byαΏ¬Ξ±ΞΌΞ¬ ΀αΟΟΞΏΟλη Modified over 6 years ago
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Sec Math II Performing Operations with Complex Numbers
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Vocabulary β1 = ? What Number multiplied by itself equals β 1? π₯ 2 = - 1 ( 1 ) ( 1 ) = 1 ( -1 ) ( - 1 ) = 1 THERE IS NO NUMBER ! Imaginary number βiβ β 1 = i π 2 = ? π 2 = ( β 1 ) Β² = - 1
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Think of the βcomplex numbersβ as the universe of numbers
Think of the βcomplex numbersβ as the universe of numbers. Complex Numbers Real Numbers Imaginary Numbers
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Complex Numbers i β1 12 2 e Real # βs Imaginary # βs Rational # βs
Whole #βs 0,1,2,3,β¦β¦β¦ β1 Integers .. -2,-1, 0, 1, 2,3 β¦β¦.. Rational #βs ( a/b) (1/2), (2/3) 5,(1/3)β¦. i e β¦.. Irrational # βs Real # βs Imaginary # βs
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What does it mean? i = β1 2 (x) = 2 x 3 (i)= 3 i ( 3i ) ( 4i ) = 3 β i β 4 β i = 3 β 4 β i β i = 12 i Β² iΒ² = ( βπ ) Β² = iΒ² = 12 ( -1) = - 12
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Simplify the Square Roots
β5 = β1 5 = π 5
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Your turn Simplify the following square roots 1. β27 β20 3. β25
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Vocabulary A Complex Number: either a real number, and imaginary number, or a combination of the 2. a + bi i Standard Form Imaginary Number Complex conjugates: two complex numbers of the form a + bi a β bi 2 + 3i β 3i
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Adding and Subtracting Complex Numbers
Real numbers and complex numbers are not βlike term.β ( 5 + 5i ) + ( 4 + 6i) 5 + 5i i Get rid of the parenthesis i + 6i Combine like terms i Treat it just like any other variable when you add them.
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Your Turn: Simplify: (add/subtract) 13. (6i) + (7i) 13a. ( 2 + 3i ) + ( 5i + 6) 13b. (7i) + ( 5 + 2i) 14. (4i) β (8i) 14a. (6i) β ( -2 β 7i) 14b. ( 6i + 4 ) β ( i)
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Multiplying Complex Numbers
1 - ( 3i) ( 4i ) = 3 * i * 4 * i = 3 * 4 * i * i = 12iΒ² = 12 ( - 1 ) = - 12 2 - ( 3 + 2i ) ( 1 β 3i ) = ?? FOIL i + 2i β 6iΒ² 3 β 7i β 6 ( -1) 3 β 7i + 6 9 β 7i
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Your turn: i ( 6i) i ( 4 + 3i ) ( 3 + 4i ) ( 8 + 9i)
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Division of complex numbers
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Your turn: π ππ π+ππ ππ π βπ ππ
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Conjugates: Your turn Simplify ( multiply the conjugates)
( x β 2 ) ( x + 2 ) ( π ) ( π ) What happens with you multiply the two conjugates together?
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Your turn: Multiply the conjugates together. ( 3 + i ) ( 3 β i )
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Same thing with complex numbers
π π βππ imaginary number not allowed π π βππ * π+ππ π+ππ = π+ππ πΒ² βπππΒ² = π+ππ π+ππ = π+ππ ππ ( π ππ + ππ ππ ) = π ππ + ππ ππ
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Your turn: 25. π π βππ 26. π π+π 27. π π βπ
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More problems ( 3 + 4i ) ( 5 β 2i ) 29. 30. π π
30. π π 31. Solve : y = 8 ( x + 3) Β² + 16 1 β2π 2+3π
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The End
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