Complex numbers 1.5 True or false: All numbers are Complex numbers

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?

POD Remember that substitution? Let’s factor with it. Ooh, a difference of cubes and a sum of cubes. How do you factor those?

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?

POD Turns out we have some imaginary solutions. Use the quadratic formula to find them.

POD All six solutions:

A brief review of i What is i? What would i 2 equal? What about i 3 or i 8 ?

A brief review of i What is i? What would i 2 equal? -1 What about i 3 or i 8 ?

A brief review of their form a + bi (real component) (imaginary component) What do you have when a = 0? What about when b = 0?

Equivalent complex numbers a + bi = c + di only if a = c and b = d Solve for x and y: (2x-4) + 9i = 8 + 3yi

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

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.

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.

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)

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.

Multiplying complex numbers (a + bi)(c + di) = (ac - bd) + (ad + bc)i Multiply: (3 + 4i)(2+5i) = i Multiply: (3 - 4i)(2+5i) = i Try this on calculators.

Complex conjugates What is the complex conjugate of a + bi? What is the product of a + bi and its complex conjugate?

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

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

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 = i i(3 + 2i) 2 = i i 51 = i 3 = -i

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?

Rational expressions with complex numbers Simplify

Complex numbers as radical expressions Multiply Hint: Rewrite using i.

Complex numbers as radical expressions Multiply. Without using i, we’d have three different radicals, and wind up with a different real number component.