Chapter 1.5 - Complex Numbers What you should learn 1.Use the imaginary unit i to write complex numbers 2.Add, subtract, and multiply complex numbers 3.

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Presentation transcript:

Chapter Complex Numbers What you should learn 1.Use the imaginary unit i to write complex numbers 2.Add, subtract, and multiply complex numbers 3. Use complex conjugates to write quotient of two complex numbers in standard form 4. Find complex solutions of quadratic equations

Imaginary unit Imaginary unit i is defined as i = √-1 where i 2 = -1

Definition If a and b are real numbers, the number a + bi is a complex number, and it is said to be written in standard form. If b = 0, the number a + bi = ais a real number If b = 0, the number a + bi is called an imaginary number.

A number of the form bi, where b = 0, is called a pure imaginary. Complex Numbers a + bi Real Numbers, a Imaginary Numbers, bi

Example: p. 127 Write the number in standard form √-8 simplify √-8 = 1 + 2√2 i 18. – 4i 2 + 2i simplify – 4i 2 = - 4 ( -1) + 2i = 4 + 2i

Equality of Complex Numbers Two complex numbers a + bi and c + di, written in standard form, are equal to each other a + bi = c + di if and only if a = c and b = d.

Operations on Complex Numbers If a + bi and c + di are two complex numbers written in standard form, their sum and difference is defined as Sum: (a + bi) + (c + di) = (a + c) + ( b+ d)I Difference: (a + bi) - (c + di) = (a - c) + ( b - d)i

Multiplication: (a + bi)(c + di) = (ac – bd) + (ad + bc) i Division: (a + bi)/ (c + di) = a + bi. c - di = c + di c - di = ac + bd + bc – ad i c 2 + d 2 c 2 + d 2

The complex conjugate The complex conjugate of a + bi is a – bi (a + bi)(a – bi ) = a 2 + b 2

The additive identity in the complex number system is zero (the same as in the real number system). Additive inverse - ( a + bi) = - a – bi So that (a + bi) + ( -a – bi) = 0 + 0i = 0

So that (a + bi) + ( -a – bi) = 0 + 0i = 0 Also, if √-a = √a i for a > 0

Example: p. 127 Perform the operation and write the result in standard form 22. (13 – 2i) + (-5 + 6i) 26. (8 + √-18) – (4 + 3√2 i)

Solution 22. (13 – 2i) + (-5 + 6i) = (13 – 5) + (-2i + 6i) = 8 + 4i 26. (8 + √-18) – (4 + 3√2 i) ( 8 + 3√2 i) + (4 + 3√2 i) = ( 8 – 4) + (3√2 i – 3√2 i) = 4 + 0i = 4

Multiplying complex numbers 1. 5 ( i) distribute 5 = i 2. ( 3 + 6i) (4 – 5i) use foil method = 3(4) - 3 (5i) + 6i(4) + (6i)(-5i) = i + 24i – 30i 2 = i + 24i = i

(5 + 3i) 2 = (5 + 3i) ( 5 + 3i) then use the foil method Try 1. 8 ( 2 + 7i) 2. (4 – 6i) ( 2 + 3i) 3. (2 – 5i) 2

Multiply by its complex conjugates i = (3 + 5i)(3 – 5i) then solve 2. (4 – 2i) = (4 – 2i) (4 + 2i) then solve

More examples Write the quotient in standard form 6 – 7i = 6 – 7i i then multiply 1 – 2i 1 – 2i 1 + 2i Solve the quadratic equation 2. 4x 2 + 6x + 5 = 0 (hint: use Quadratic Formula)

Try Solve 3. Watch video Watch video from Cengage Learning (webassign) 3 x 2 – 2x + 5 = 0