Lesson A.6: Complex Numbers; Quadratic Equations in the Complex Number System
Guided Notes
Intro to Complex Numbers
We have discussed how the square of a real number cannot be negative. Therefore, there is no real number x
for which x 2 1. Thus, we introduce the _______________ denoted by i and whose square is -1.
i 1
i 2 ( 1) 2 ___
i 3 i 2 i 1 i ___
i 4 i 2 i 2 1 1 ___
Ex. i 9 i 4 i 4 i1 11 i i
Practice: What does i19
Complex numbers are numbers of the form _________where a and b are real numbers. The real number
a is called the ___________
, and the real number b is called the _________________.
Adding, Subtracting, and Multiplying Complex Numbers
To add complex numbers, simply add the real parts together and add the imaginary parts together.
(a bi) (c di) __________________
Ex. (2 3i) (1 4i) (2 1) (3 4)i 3 i
The same idea holds true for subtracting complex numbers.
(a bi) (c di) __________________
Ex. (3 i) (4 3i) (3 4) (1 3)i 1 4i
To multiply complex numbers, follow the usually rules for multiplying binomials and remember i 2 1.
(a bi) (c di) ____________________
Ex. (1 3i) (4 5i) 1(4 5i) 3i(4 5i) 4 5i 12i 15i 2 4 7i 15 19 7i
Practice: Given the following complex numbers, perform in the indicated operations.
z 1 4i
w 5 3i
1) w z
2) w z
3) w z
double distribution
Conjugates of Complex Numbers
If z a bi is a complex number, then its ______________, denoted z , is defined as
z a bi _________
Ex. If z 2 3i then its conjugate is z 2 3i
Let’s see what happens if we multiply
a complex numbers with its conjugate!
zz ________
abi b2i 2 a2 b2
zz (a bi)(a bi) a2 abi
Ex. (2 3i)(2 3i) 4 6i 6i 9i 2 4 9 13 Or using formula (2 3i)(2 3i) 22 32 4 9 13
Practice. Find the conjugate of the following complex numbers and then multiply.
1) z 3 4i
2) z 1 8i
3) z 2i
Dividing Complex Numbers
divide complex numbers, we can multiply the numerator and denominator by the conjugate of the
To
denominator .
Ex.
2 3i ______
2
4 3i
4 (3) 2 16 9 25 25 25
Practice:
3 i
2 4i
Let’s Practice What We’ve Learned!
Given the following complex numbers, perform the indicated operations. w 2 3i
1) w z
2) w z
3) w z
4)
w
z
z 5 2i
Solving Quadratic Equations with a Negative Discriminant
We have discussed how a quadratic equation with a negative discriminant has no real number solution. For
example,
Cannot square root a negative in
Ex 1: x 2 4 0
x 2 4
x 4
the real number system
However,
if we extend our
number system to allow complex numbers, quadratic equations will __________
_______________. Thus, if N is a positive real number, we define the __________________,
denoted N , as
N ______
Ex 2.
Ex 3.
Ex 4.
1 1i i
4 4i 2i
8 8i 2 2i
Practice 1:
Practice 2:
3
18
Let’s apply this to solving quadratic equations in the complex number system.
Ex 5. Solve x 2 4 0
x2 4 0
x 2 4
x 4
x 2i
x 4i
Practice
3: Solve x 2 9 0
Now let’s look
at quadratic equations where we must complete the square to solve.
Ex 6. Solve x 2 4 x 8 0 in the complex number system
x 2 4 x 4 8 4
(x 2) 2 4
x 2 4
x 2 2i
x 2 2i
Practice 4: Solve x 2 2x 4 0 in the complex number system
Character of the Solutions of a Quadratic Equation
In the complex number system, consider a quadratic equation ax 2 bx c 0 with real coefficients.
1. If b 2 4ac 0 , the equation ___________________________________.
2. If b 2 4ac 0 , the equation ___________________________________
.
3. If b 2 4ac 0 , the equation ___________________________________.
Ex 7. Given x 2 4 x 8 from example 6, we can determine the character of its solution(s)
b 2 4ac (4)2 4(1)(8) 16 0
Thus, two complex solutions
This matches our solution of x 2 2i from example 6.
Ex 8. Determine the character of the solutions for 9x 2 6x 1 0 , then solve for x.
Thus, repeated real root
b 2 4ac (6)2 4(9)(1) 0
We can factor to solve the equation
9x 2 6x 1 0
(3x 1)(3x 1) 0
x 13 , x 13
Repeated real root!
Ex 9. Determine the character of the solutions for x 2 4 x 1 0 , then solve for x.
Thus, two unequal real solutions
b2 4ac (4)2 4(1)(1) 12 0
We can complete the square to solve
the equation
x 2 4x 4 1 4
(x 2)2 3
x 2 3
x 2 3
Two unequal real solutions!
Practice 5: Determine the character of the solutions for x 2 6x 13 0, then solve for x.
0
