Objectives for Class 3 Add, Subtract, Multiply, and Divide Complex Numbers. Solve Quadratic Equations in the Complex Number System

Basics of the Complex Number System (Square root of -1) = i Simplify the following square roots (Square root of -25) Pull the square root of -1 out first to form i(square root of 25) Break down the square root of 25 (square root of -25) = 5i

(square root of -81) 9i (square root of -20) 2i(square root of 5) Simplify each of the following. Hit “enter” to check your answers. See Mrs. Dorshorst for help if needed. (square root of -81) 9i (square root of -20) 2i(square root of 5)

Complex Numbers: a + bi ‘a’ is called the real part ‘b’ is called the imaginary part Standard Form: a + bi Ex: 3 – 5i Real Part is 3 Imaginary part is -5i Real number: a Ex: 8 No imaginary part Pure imaginary: bi Ex: 3i No real part

Equal Complex Numbers: a + bi = c + di a must equal c : Real parts must be equal b must equal d: Imaginary parts must be equal Ex: 3x + 5yi = -7 – 4i Set real parts equal and solve for variable:3x =-7 X = -7/3 Set imaginary parts equal and solve for variable: 5yi = -4i 5y = -4 so y = -4/5

8x – 9yi = 4 + 2i 8x = 4 x = 1/2 -9y = 2 y = -2/9

Addition and Subtraction of Complex Numbers Sum of Complex Numbers: (a + bi) + (c + di) = (a + c) + (b + d)I Ex: (4 – 7i) + (12 + 19i) (4 + 12) + (-7i + 19i) 16 + 12i Try: (8 + 52i) + (-12 – 40i) Solution: -4 + 12i

Difference of Complex Numbers: (a + bi) – (c + di) = (a – c) + (b – d)i Ex: (-3 + 7i) – (-8 + 9i) Distribute the subtraction through the second complex number (-3 + 7i) + ( +8 – 9i) Add like terms (-3 + 8) + (7i – 9i) 5 – 2i

Try: (25 – 3i) – (13 – 10i) (25 – 13) + (-3i + 10i) 12 + 7i

Examples (2 – 5i) – (8 + 6i) Solution: -6 – 11i (3 + 5i) + (6 – 8i)

Product of Complex Numbers Distribute each term through the next parenthesis. Remember that i2 = -1 Examples (5 + 3i)(2 + 7i) 10 + 35i + 6i + 21i2 10 + 35i + 6i – 21 -11 + 41i

(3 – 4i)(2 + i) 6 + 3i – 8i -4i2 6 + 3i – 8i + 4 10 – 5i 3i(-3 + 4i) -9i - 12

Conjugate Same values with opposite middle sign Ex: 2 + 3i conjugate: 2 – 3i _______ Notation: a + bi ____ Find the conjugate of 3 + 4i 3 – 4i

Multiply 4 – 2i times its conjugate (4 – 2i)(4 + 2i) Conjugate is 4 + 2i) 16 + 8i – 8i - 4i2 Distribute 1st parenthesis through second parenthesis 16 + 4 Combine like terms and substitute -1 in for i2 20

Multiply 3 + 2i times its conjugate What is the conjugate? 3 – 2i Do the multiplication. (3 + 2i)(3 – 2i) Solution: 13

Dividing Complex Numbers Multiply both the numerator and the denominator times the conjugate to eliminate the radical in the denominator (recall that ‘i’ is a radical) Example: (1 + 4i) / (5 – 12i) ((1+4i)(5+12i))/((5-12i)(5+12i)) Multiply by conjugate of denominator (5 + 12i + 20i +48i2)/(25 + 60i – 60i – 144i2) Multiply numerators and multiply denominators (5 + 32i – 48)/(25 + 144) Simplify (-43 + 32i) / (169)

(2 – 3i) / (4 – 3i) What would you multiply by to eliminate the radical in the denominator? 4 + 3i Multiply numerator and multiply denominator ((2 – 3i)(4 + 3i))/((4 – 3i)(4+ 3i)) Simplify and find answer Solution: (17 – 6i) / (25)

Other Examples If z = 2 – 3i and w = 5 + 2i find each of the following _____ Find (z + w) What does this notation mean? Conjugate of the sum Add then state conjugate. 7 – 1i so answer is 7 + 1i

Powers of i Recall that i2 = -1 To find high level powers of i: Ex: i35 1.Break down the expression into i2 notation. (i2)17i Substitute -1 in for i2 (-1)17i Simplify as far as possible. (-1)i

Simplify: i7 + i25 – i8 Simplify each power of i Combine like terms i7 = (i2)3i = (-1)3i = -1i i25 = (i2)12i = (-1)12i = 1i i8 = (i2)4 = (-1)4 = 1 Combine like terms -1i + 1i – (1) = 1

Solving Quadratic Equations in the Complex Number System Solve: x2 – 4x = -8 1. Put into standard form. x2 – 4x + 8 = 0 2. Solve using the Quadratic Formula. When simplifying the radical remember that the square root of -1 is i. (4 + sqrt(16 – 4(1)(8))) / 2 (4 + sqrt -16)/2 (4 + 4i)/2 2 + 2i

Discriminant: b2 – 4ac Value under the radical in the quadratic formula is called the discriminant. The value of the discriminant determines the type of solution for the equation. D > 0: discriminant is positive there are 2 real solutions D < 0: discriminant is negative there are 2 complex solutions D = 0: discriminant is o there is one real solution (double root)

What type of solutions do the following equations have What type of solutions do the following equations have? Do not solve just determine the type of solutions. (Recall the rules for discriminants on the previous slide. 3x2 – 8x = 9 b2 – 4ac => 64 – 4(3)(-9) D > 0 2 real number solutions 4x2 + 3 = 5x b2 – 4ac => 25 – 4(4)(3) D > 0 2 complex number solutions

Assignment Page 117 #9, 13, 19, 23, 25, 27, 31, 35, 39, 49, 51, 55, 59, 63, 75, 77, 83,