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Published byAsher Phillips Modified over 9 years ago
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Complex numbers 1.5 True or false: All numbers are Complex numbers
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POD Consider our final equation from the previous lesson. We determined that x = -2 and x = 1 were solutions. How many solutions will there be, counting multiplicities and complex solutions? Are -2 and 1 the only solutions, or are there other solutions as well? How could we find out?
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POD Remember that substitution? Let’s factor with it. Ooh, a difference of cubes and a sum of cubes. How do you factor those?
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POD Let’s do a complete factorization. Finding all the factors helps us find all the solutions. We can see our two solutions easily. Do they have a multiplicity greater than one? (CAS does this easily.) How do we find the other solutions?
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POD Turns out we have some imaginary solutions. Use the quadratic formula to find them.
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POD All six solutions:
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A brief review of i What is i? What would i 2 equal? What about i 3 or i 8 ?
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A brief review of i What is i? What would i 2 equal? -1 What about i 3 or i 8 ?
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A brief review of their form a + bi (real component) (imaginary component) What do you have when a = 0? What about when b = 0?
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Equivalent complex numbers a + bi = c + di only if a = c and b = d Solve for x and y: (2x-4) + 9i = 8 + 3yi
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Equivalent complex numbers a + bi = c + di only if a = c and b = d Solve for x and y: (2x-4) + 9i = 8 + 3yi 2x-4 = 89 = 3y x = 6y = 3
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Adding complex numbers (a + bi) + (c + di) = (a + c) + (b + d)i Add: (3 + 4i) + (2 - 5i) Subtract: (3 + 4i) - (2 - 5i) Try this on calculators.
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Adding complex numbers (a + bi) + (c + di) = (a + c) + (b + d)i Add: (3 + 4i) + (2 - 5i) = 5 - i Subtract: (3 + 4i) - (2 - 5i) = 1 + 9i Try this on calculators.
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This is how it looks on the TI-84. (a + bi) + (c + di) = (a + c) + (b + d)i Add: (3 + 4i) + (2 - 5i) Subtract: (3 + 4i) - (2 - 5i)
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Multiplying complex numbers (a + bi)(c + di) = (ac - bd) + (ad + bc)i Multiply: (3 + 4i)(2+5i) Multiply: (3 - 4i)(2+5i) Try this on calculators.
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Multiplying complex numbers (a + bi)(c + di) = (ac - bd) + (ad + bc)i Multiply: (3 + 4i)(2+5i) = -14 + 23i Multiply: (3 - 4i)(2+5i) = 26 + 7i Try this on calculators.
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Complex conjugates What is the complex conjugate of a + bi? What is the product of a + bi and its complex conjugate?
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Complex conjugates What is the complex conjugate of a + bi? What is the product of a + bi and its complex conjugate? That means the factorization for is
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Operations with complex numbers Express in a + bi form: 4(2 + 5i) - (3 - 4i) (4 - 3i)(2 + i) (3 - 2i) 2 i(3 + 2i) 2 i 51
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Operations with complex numbers Express in a + bi form: 4(2 + 5i) - (3 - 4i) = 5 +24i (4 - 3i)(2 + i) = 11 – 2i (3 - 2i) 2 = 5 - 12i i(3 + 2i) 2 = -12 + 5i i 51 = i 3 = -i
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Rational expressions with complex numbers Simplify Hint: What is the complex conjugate of the denominator? And try this on calculators. How do you get rational coefficients?
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Rational expressions with complex numbers Simplify
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Complex numbers as radical expressions Multiply Hint: Rewrite using i.
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Complex numbers as radical expressions Multiply. Without using i, we’d have three different radicals, and wind up with a different real number component.
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