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Sect P.6 … Complex Numbers. The Solutions to x 2 + 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. - PowerPoint PPT Presentation
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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
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