1. Chapter 5, Section 9 Complex Numbers

2. i, the imaginary number By definition, the square root of -1 is defined as i, i.e. We can now write the square root of a negative number, let’s say, -9, this way: It is easy to show that if i is the square root of negative 1, then i 2 should be -1. i 3 is just i 2 times i, so it is -1i i 4 is just i 2 times i 2, so it is 1, the same as i 0 For powers higher than 4, divide the exponent by four and keep the remainder (also called the “modulo” operation); e.g. i 19 = i 3 = -i.

3. Example 1 Simplify We will factor this out, separating the negative sign: Most textbooks prefer putting the i in FRONT of the radical, so it doesn’t look like it’s UNDER the radical, but you may prefer to put it at the end, like any other variable. or

4. Example 2 Simplify We will again factor this completely, separating perfect squares:

5. Example 3 Simplify -3i · 2i We will use the commutative and associative properties: -3i · 2i -3 · 2 · i · i -6 · i 2 -6 · -1 6

6. Example 4 Simplify Simplify i 35 We will remember to divide the exponent by 4 and keep the remainder: i 35 = i 3 = -i

7. Example 5 Solve 5y 2 + 20 = 0 Original problem Subtract 20 from both sides Divide both sides by 5 Take square root of both sides Simplify. Yes, the book is inconsistent!

8. Complex numbers Complex numbers have a real AND an imaginary component. For any two complex numbers to be equal, their real parts must be equal, and their imaginary parts must be equal. I.e. a + bi = c + di only if a = c and b = d

9. Example 6 Find the values of x and y that make the equation true: 2x + yi = -14 - 3i We will remember that for two complex numbers to be equal, the real components must be equal, as must the imaginary components: 2x = -14yi = -3i x = -7 y = -3

10. Complex number arithmetic When adding and subtracting complex numbers, add or subtract the real components then add or subtract the imaginary component. When multiplying, FOIL, and remember that the value of i 2 = -1. When dividing, do NOT leave an imaginary component in the denominator; multiply by i to rationalize a purely imaginary number, and multiply a complex number by its COMPLEX conjugate. The complex conjugate of a + bi is a - bi.

11. Example 7 Simplify (3 + 5i) + (2 - 4i) We will use the commutative and associative properties -- AKA “combine like terms”: (3 + 5i) + (2 - 4i) 3 + 5i + 2 - 4i 3 + 2 + 5i - 4i 5 + i

12. Example 8 Simplify (4 - 6i) - (3 - 7i) Remember to distribute the -1 over the second set of parentheses! (4 - 6i) - (3 - 7i) 4 - 6i - 3 + 7i 4 - 3 - 6i + 7i 1 + 1i 1 + i

13. Example 9 Simplify (1 + 4i) · (3 - 6i) We need to remember to FOIL (use the Distributive Property): (1 + 4i)(3 - 6i) 3 - 6i + 12i - 24i 2 3 + 6i - 24(-1) 3 + 6i + 24 27 + 6i

14. Example 10 Simplify Because the denominator is purely imaginary, we will multiply by i over i:

15. Example 11 Simplify We will recognize that the denominator is COMPLEX, and we will multiply numerator and denominator by the complex conjugate:

16. Your assignment… Chapter 5 section 9, 19-63 and 71-85 odd, PLUS 28, 44, 54, 64, and 78 (for 64, in the formula E = I·Z, E is voltage, I is current, and Z is impedence). “What we obtain too cheap, we esteem too lightly: it is dearness only that gives every thing its value.” --Thomas Paine