Assignment, red pen, highlighter, textbook, GP notebook

1. total: Assignment, red pen, highlighter, textbook, GP notebook M3D9 Have out: Bellwork: Simplify: 3) 2) 1) +1 +1 20i2 –3i +2 12i2 –20 +2 –12 +2 5) 4) .

2. Complex Numbers, Part 2 Recall from yesterday: When a complex number is simplified, it must be written in _______ form, such that a is the ____ ____ and b is the ________ ____. real part a + bi part imaginary real four imaginary two two Example: Example: 2 – 6 – 4i + 10i 15 – 3i + 5i – i2 –4 + 6i – (–1) 15 + 2i 16 + 2i

3. Dividing Complex Numbers Dividing complex numbers requires some creativity. For example: = ??? Recall simplifying expressions such as . rationalize To “_________” the denominator, we multiply the numerator and denominator by _____. Then = To simplify complex quotients, we need to make the denominator ______. To do this, we multiply top and bottom by the _________ _________ of the denominator. real complex conjugate Definition: For a + bi, the _______ _________ is _____. complex conjugate a – bi

4. How does the complex conjugate work? Multiply (a + bi)(a – bi). a + bi a + abi a2 – bi – abi – b2i2 –b2(–1) (–abi) (a + bi)(a – bi) = ___ + ___ + ____ + ____ a2 abi (–b2i2) a2 + b2 = __________ real The product of a complex number and its conjugate is ____ and is _______. a2 + b2

5. 2 – i Back to the example . The conjugate of is ______. (2 – i) 5 – 5i (2 – i) 5 3 – i 2 + i 2 2 – 2i + 2i 6 4 – i – 3i + i2 – i – 2i – i2 6 – 2i – 3i – 1 4 – 2i +2i + 1 5 – 5i 5

6. Practice: Simplify. Write all answers in a + bi form. (3 – 2i) This is not a + bi form. We must ALWAYS write the answer as 2 terms (or fractions). 6 – 4i (3 – 2i) 13 9 – 6i + 6i – 4i2 9 + 4 4i + i2 (4 + i) –1 + 4i (4 + i) 17 16 – 4i + 4i – i2 16 + 1

7. Practice: 6 + 15i + 2i + 5i2 6 + 17i – 5 (2 + 5i) 1 + 17i (2 + 5i) 29 4 + 10i – 10i – 25i2 4 + 25

8. The Powers of i So what is i? Moreover, what are the powers of i? Let’s start at the beginning… anything raised to the zero power is 1. (yeah, yeah, except zero to the zero power) Therefore i0 = 1 We also know that i1 = by the definition of i. What is i2? What about i3,i4, . . . ?

9. Recall the complex plane from yesterday. Label one unit on each axis. 1 i0 = _____ i i1 = _____ Imaginary –1 i2 = _____ (i2)(i) (–1)(i) –i i3 = _____ = _____ = ____ i1 = i i (i2)(i2) (–1)(–1) 1 i4 = _____ = _____ = ____ i0 = 1 1 Real (i4)(i) (1)(i) i i5 = _____ = _____ = ____ i4 = 1 i2 = –1 –1 –i i3 = –i (i4)(i2) (1)(–1) –1 i6 = _____ = _____ = ____ (i4)(i3) (1)(–i) –i i7 = _____ = _____ = ____ Notice that the units on the plane match our answers for the powers of i. Label the first 4 or so powers on the axis. (i4)(i4) (1)(1) 1 i8 = _____ = _____ = ____ i9 = _____ i –1 i10 = _____

10. cycle four The powers of i are in a ______ with _____ elements: ___, ___, ___, ___. i –1 –i 1 IM i0 = 1 To compute a power of i, start at _____, then go around the circle ________________. i1 = i counterclockwise i0 = 1 jumps The number of ______ is equal to the _______. RE i2 = –1 power 25 i102 will go around the cycle ___ times and land on ___ = ____. i3 = –i i2 –1 i3 –i i47 = ___ = ___. 4 remainder Just divide the power by ___ and look at the _________. i1 i 39 i157 = ____ = ____. R:1

11. Another example: 3 R:3 i–15 i–3 i Negative power means “go the opposite direction” IM So: i1 = i i0 = 1 remainder = 1 i1 RE i2 = –1 remainder = 2 i2 i3 = –i remainder = 3 i3 remainder = 0 (no remainder) i4 Of course, reduce these if you can.

12. 1. i53 i1 i IM 2. i176 i0 1 i1 = i 3. i–6 i–2 –1 i0 = 1 4. i–11 i–3 i RE i2 = –1 i3 = –i

13. Finish the worksheets.