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
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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  11 i  i
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Practice: What does i19 
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Complex numbers are numbers of the form _________where a and b are real numbers. The real number
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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
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The same idea holds true for subtracting complex numbers.
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(a  bi)  (c  di)  __________________
Ex. (3 i)  (4  3i)  (3 4)  (1 3)i  1 4i
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To multiply complex numbers, follow the usually rules for multiplying binomials and remember i 2  1.
(a  bi)  (c  di)  ____________________
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Ex. (1 3i)  (4  5i)  1(4  5i)  3i(4  5i)  4  5i 12i 15i 2  4  7i 15  19  7i
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Practice: Given the following complex numbers, perform in the indicated operations.
z  1 4i
w  5  3i
1) w  z
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2) w  z
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3) w  z
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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
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Let’s see what happens if we multiply
 a complex numbers with its conjugate!
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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
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Practice. Find the conjugate of the following complex numbers and then multiply.
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1) z  3 4i
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2) z  1 8i
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3) z  2i
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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
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1) w  z
2) w  z
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3) w  z
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4)
w
z
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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 __________
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_______________. Thus, if N is a positive real number, we define the __________________,
denoted N , as
N  ______
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Ex 2.
Ex 3.
Ex 4.
1  1i  i
4  4i  2i
8  8i  2 2i
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 Practice 1:
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Practice 2:
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3 
18 
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Let’s apply this to solving quadratic equations in the complex number system.
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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
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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
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x  2   4
x  2  2i
x  2  2i
Practice 4: Solve x 2  2x  4  0 in the complex number system
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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 ___________________________________.
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2. If b 2  4ac  0 , the equation ___________________________________
.
3. If b 2  4ac  0 , the equation ___________________________________.
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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
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Thus, two complex solutions
This matches our solution of x  2  2i from example 6.
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Ex 8. Determine the character of the solutions for 9x 2  6x  1  0 , then solve for x.
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Thus, repeated real root
b 2  4ac  (6)2  4(9)(1)  0
We can factor to solve the equation
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9x 2  6x  1  0
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(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.
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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
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(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.
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