Sect P.6 … Complex Numbers

Sect P.6 … Complex Numbers The Solutions to x2 + 9 = 0

Definition… • A complex number … any number written in the form a + bi • a is the real part • b is the imaginary part • a+ bi is the Standard Form.

Things to remember… • All real numbers are also Complex Numbers. • a can be 0, so 6i is an imaginary numbers • Two complex numbers are equal iff …

Adding Complex numbers • Add the “real” parts together and add the “imaginary” parts together. • (7 + 3i) – (4 + 5i) = • Now do Exercise 3, p.52

Additive Identities… • The additive Identity of a + bi is –(a + bi) = – a – b Because a + bi + (– a – bi) = 0

Multiplying Complex Numbers • Method 1: Use the Distributive Property… (2+3i)(5 – i ) = 2(5 – 1 ) + 3i(5 – i ) • Or, FOIL Method… (2+3i)(5 – i ) =

Raising a Complex Number to a Power • (2 + 3i)2 = (2+3i)(2+3i)=

Complex Conjugates • If z = a + bi is a complex number, • Then, the COMPLEX CONJUGATE of z is z = a – bi And (a + bi)(a – bi ) = a2 – b2

Dividing Complex numbers… • Multiply by 1 …in the form • Example:

Complex Solutions of Quadratic Equations • When using the Quadratic Formula… • If the radicand is <0 (negative), then there is no real solution… because the square root of a negative number is not a “real” number. • The radicand b2 – 4ac is called the Discriminant.

Discriminant of Quadratic Equations • For all equations of the form ax2 +bx + c = 0 • If b2 – 4ac >0, then there are two distinct solutions • If b2 – 4ac = 0, there is one repeated solution • If b2 – 4ac <0 , then there is a complex conjugate pair of solutions.

Example… • Solve Algebraically… x2 + x + 1 = 0 using the Quadratic Formula • X =

Homework • P. 52 …. • Quick Review…. 2, 8 • Exercises … 4, 10, 18, 26, 34, 36, 42, 48

Sect P.6 … Complex Numbers