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

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**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

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**(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)

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**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: No imaginary part Pure imaginary: bi Ex: 3i No real part

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**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 = so y = -4/5

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

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**Addition and Subtraction of Complex Numbers**

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

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**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

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

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

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**Product of Complex Numbers**

Distribute each term through the next parenthesis. Remember that i2 = -1 Examples (5 + 3i)(2 + 7i) i + 6i + 21i2 i + 6i – 21 i

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

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**Conjugate Same values with opposite middle sign**

Ex: i conjugate: 2 – 3i _______ Notation: a + bi ____ Find the conjugate of i 3 – 4i

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**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

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**Multiply 3 + 2i times its conjugate**

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

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**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)/( i – 60i – 144i2) Multiply numerators and multiply denominators (5 + 32i – 48)/( ) Simplify ( i) / (169)

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(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)

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**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

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**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

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**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

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**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

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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)

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**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 => – 4(3)(-9) D > real number solutions 4x2 + 3 = 5x b2 – 4ac => – 4(4)(3) D > complex number solutions

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

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