Bell Ringer
Multiplying and Dividing with Complex Numbers Monday, February 29, 2016
Multiplying Complex Numbers When we multiply complex numbers, sometimes the imaginary part “disappears”. Ex: (2 + 3i) (2 – 3i) = (4 – 6i + 6i – 9i 2 ) = (4 + 9) = 15
Theorems say… Complex roots come in pairs… Ex: You can’t have a root of 3i without a root of -3i
Multiplying Complex Numbers If a polynomial has complex roots, when you multiply the factors to find the original polynomial, the imaginary part “disappears”.
What about dividing complex numbers? You can’t divide by an imaginary number. To make the denominator real, we must multiply by a special 1 called the CONJUGATE. We make the middle term “disappear”. 6 – 3i ÷ 2 + 4i 6 – 3i (2 – 4i) 2 + 4i (2 – 4i) 12 – 30i + 12i 2 4 – 16i 2 12 – 30i + (-12) 4 – (-16) - 30i 20 -3 i 2
Complex numbers and Conjugates Complex numbers are in the form a + bi Complex conjugates are in the form a - bi Complex: 2 + 4i Conjugate: 2 – 4i Complex: i Conjugate: -3 – 5i Complex: 1 – 7i Conjugate: 1 + 7i
Time to Practice! Classwork: Find Polynomial Functions given the roots (multiplying) Homework: Dividing Complex Numbers (multiplying by i and complex conjugates)
Exit Ticket 1. What happens when you multiply complex conjugates? 2. Both complex and irrational roots must come in _____. 3. How do you make the denominator real if it is a complex number?