Complex Numbers Using Complex Conjugates in dividing complex numbers and factoring quadratics -- Week 15 11/19

Basics of Complex Numbers Every number is complex! (could have zero for either real or imaginary part) Take the form a + bi where a is the real part i = −1 and bi is the imaginary part a + bi is a VALUE not a binomial expression, though it often acts like one. Add or Subtract by adding or subtracting real parts, then imaginary parts, and combining what you get. i.e. (3 + 2i) + (5 + i) is 8 + 3i just WATCH SIGNS!

Multiplying Complex Numbers To multiply two complex numbers, you treat them like binomials and F.O.I.L. them. This means the i2 part will become real since i2 = -1 (3 + 2i) ( 5 + i) = 15 + 3i + 10i + 2 i2 = 15+ 13i + 2 i2 = 15+ 13i -2 = 13 + 13i

Dividing Complex Numbers To DIVIDE complex number, you must simplify by getting the imaginary stuff out of the denominator. Any complex number can be made real by multiplying by its complex conjugate Complex conjugate is the same complex number with inverse sign of imaginary part: i.e. for 3 + 2i it is 3 – 2i for -4i it is 4i etc. Multiply NUMERATOR and DENOMINATOR by complex conjugate of the DENOMINATOR (3 + 2𝑖) 5+𝑖 multiply by conjugate of bottom (5 – i) (3+2𝑖) 5+𝑖 ∙ 5 −𝑖 5 −𝑖 = 15+10𝑖 −3𝑖 −2 𝑖 2 25+5𝑖 −5𝑖 − 𝑖 2 Simplify: 17+7𝑖 26

Factoring to Complex Conjugates Now that we have complex numbers, we can factor sums of squares!!! X2 + 9 means X2 = -9 so x = ± −9 = ± 3i So factors are (x + 3i) (x – 3i) Likewise, if you are given roots (solutions) which are imaginary, you can work backward to solve for the original quadratic. Given Roots: 4i, -4i the factors would be (x + 4i) (x – 4i) so the quadratic would be x2 + 16 since x2 = -16 would give 4i, -4i

Work Together p. 254 # 47-61 odd