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Before we do anything else, put your computers away

Before we do anything else, put your computers away. Imaginary Numbers. A beginning…. Before today, you couldn’t do the square root of a negative number. Today, that will change.

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Before we do anything else, put your computers away

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  1. Before we do anything else, put your computers away

  2. Imaginary Numbers

  3. A beginning… Before today, you couldn’t do the square root of a negative number. Today, that will change. In the beginning, mathematicians saw no need for these numbers outside of the classroom, so they game them the name “imaginary numbers”

  4. However, electronic and other scientific applications have come to use them frequently. Unfortunately, the name, “Imaginary Numbers” has remained.

  5. Foundation The basis of the imaginary numbers is the square root of -1. It is given the letter i. i = √-1 So √-25 = √25 x √-1 = 5i

  6. Basically, Any time you see a negative under a square root, you pull its “i” out. You do this before you do anything else. √-36 x √-9 = 6i x 3i = 18i2

  7. Which leads to another problem… What in the world are we going to do with i2? If i = √-1, then i2 = -1. Also, i3 = -i and i4 = 1.

  8. Adding and Subtracting (3 + 4i ) + (-2 + 6i ) = (3 + -2) + ( 4i + 6i) = 1 + 10i Just add the “real” parts and add the “imaginary” parts separately.

  9. (-5 + 2i ) – ( 2 – 6i ) = (-5 + 2i ) + ( -2 + 6i ) When you’re subtracting, change the subtraction to addition by changing all the signs following the subtraction…just like we did before with parentheses!! Then combine like terms. = -7 + 8i

  10. Multiplying 3i x 4i = 12i2 (remember that i2 = -1) = -12 Isn’t that easy? 4i ( -3 + 2i ) Just distribute!! -12i + 8i2 (remember that i2 = -1) = -12i -8 and we usually write the real number first so we would put -8 – 12i

  11. More Multiplying!!! (-2 – 5i) (4 + 3i ) Just FOIL!!! -8 – 6i – 20i – 15i2 (remember that i2 = -1) -8 – 6i – 20i + 15combine like terms = 7 – 26i isn’t that easy?

  12. Just one more multiplying (3 – 5i)2 = (3 – 5i) (3 – 5i) = 9 -15i – 15i + 25i2 = 9 -15i – 15i – 25 = -16 – 30i

  13. Well…maybe one more Do you remember how multiplying conjugates got rid of the radicals? The same thing works here. Multiplying conjugates gets rid of i. ( -3 + 2i ) ( -3 – 2i ) = 9 + 6i – 6i – 4i2 = 9 + 6i – 6i + 4 = 13

  14. Now its up to you Your assignment is 5-9    pg 314    20, 22-24, 26-28, 31-33, 38- 42, 44-46, 48, 49

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