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

i2 = -1(Square root of -1) = iSimplify 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

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 = 4x = 1/2

-9y = 2y = -2/9

Addition and Subtraction of Complex Numbers

Sum of Complex Numbers: (a + bi) + (c + di) = (a + c) + (b + d)IEx: (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)Solution: 9 – 3i

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 + 410 – 5i3i(-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 + 4i3 – 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 + 4Combine 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 answerSolution: (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 sumAdd 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

2. Substitute -1 in for i2 (-1)17i

3. Simplify as far as possible. (-1)i

Simplify: i7 + i25 – i8

Simplify each power of i 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? 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,