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:

a bi

a bi

2

3 i

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.

2 4

2

b b ac

a

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 = 2 4

2

b b ac

a

HOMEWORK

P. 52 ….

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