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Complex Numbers and Equation Solving
1. Simple Equations
2. Compound Equations
3. Systems of Equations
4. Quadratic Equations
5. Determining Quadratic Equations
Complex Numbers and Equation Solving
In this section we will consider the following:
a) simple equations
b) compound equations
c) systems of equations
d) quadratic equations
Simple Equations:
Goal: isolate the variable on either the left or right side of the equation and to solve for one
positive variable.
Example 1: This example reviews the basic rules for solving a simple equation
3x  4  5(x  2)  3
3x  4  5x  10  3
1. Remove brackets or parentheses that exist within the equation
using the distributive property. Reminder: make sure the value
outside the ( ) or [ ] is distributed to all terms within, with a
3x  5x  4  10  3
special focus on any sign changes.
2. Isolate the terms containing the variable remembering that if
2x  17
terms are moved across the equal sign, the sign of the term
1
1
must change.
  x   17
3. Simplify both sides of the equation.
2
2
17
17  4. Solve for “1” positive variable by multiplying both sides by

the reciprocal of the coefficient or in simple terms divide both
x   , SS  
2
 2 
sides by the coefficient of the variable
Example 2: Real coefficient assigned to the variable with real and imaginary numbers
3x  4  5i  8x  7  2i
3x  8x  4  5i  7  2i
11x  3  3i
3  3i
3  3i 

x
, SS 

11
 11 
1. Isolate the terms containing the variable “x” on the
left side of the equation and move all real and
imaginary numbers to the right side.
2. Simplify both sides of the equation.
3. Solve for one positive “x”
Example 3: Imaginary coefficients assigned to the variable with imaginary numbers
4ix  2i  3(3ix  5i)
4ix  2i  9ix  15i
4ix  9ix  2i  15i
5ix  17i
x
17
17 
, SS 
 
5
 5 
1. Remove any parentheses.
2. Isolate the terms containing the variable “x” on the
left side of the equation and move all imaginary
numbers to the right side.
3. Simplify both sides of the equation.
4. Solve for one positive “x”. Dividing both sides of
the equation by “-5i” results in a situation where an
imaginary number is contained in both numerator
and denominator and as a result the value “i”
cancels.
Example 4: Imaginary coefficients assigned to the variable with real and imaginary numbers
4ix  3  2(ix  5i)  7  4i
4ix  3  2ix  10i  7  4i
1. Remove any parentheses.
2. Isolate terms containing the variable “x” on the left
side and the real and imaginary numbers on the
4ix  2ix  3  10i  7  4i
right side of the equation.
3. Simplify both sides of the equation
2ix  4  14i
4. When solving for one positive “x” you must divide
4  14i i 4i  14i 2
by an “i” and following the rule of no imaginary
x
 
2
2i
i
2i
number in the denominator, both numerator and
denominator must be multiplied by an “i”.
4i  14
x
 2i  7 SS 2i  7 5. Simplify the resulting fraction
2
Example 5: Real and imaginary coefficients assigned to the variable with real numbers
1. Isolate and simplify as in previous examples
2. Remove the common factor “x” from the left side of the
equation resulting in a complex number factor (7 - 2i).
7x  2ix  11
3. When solving for one positive “x” you must divide by the
complex number factor and following the rule of no
x(7  2i)  11
imaginary number in the denominator, the denominator
11
(7  2i) 77  22i
and numerator on the right side must be multiplied by
x


2
(7  2i) (7  2i) 49  4i
the conjugate of the complex number factor
77  22i
77  22i
77  22i  (7 + 2i).
x

SS 


49  4(1)
53
 53  4. Simplify the resulting fraction
5x  7  2ix  4ix  4  2x
5x  2ix  4ix  2x  7  4
Example 6: Real and imaginary coefficients assigned to the variable with imaginary numbers
4ix  5i  2x  3(2ix  3i)  4x
4ix  5i  2x  6ix  9i  4x
4ix  2x  6ix  4x  5i  9i
2ix  6x  14i
x(2i  6)  14i
14i
(2i  6) 28i 2  84i
x


2
(2i  6) (2i  6)
4i  36
28(1)  84i 28  84i
x

4(1)  36
40
7  21i 
SS 

 10 
1. Remove any parenthesis.
2. Isolate the variables on the left side and
the imaginary numbers on the right side of
the equation.
3. After simplifying the equation, remove the
common factor “x” from the left side of
the equation resulting in a complex
number factor (-2i - 6).
4. When solving for one positive “x” you
must divide by the complex number factor
and following the rule of no imaginary
number in the denominator, the
denominator and numerator on the right
side must be multiplied by (-2i + 6).
5. Simplify the resulting fraction. Some
basic items to take into consideration:
a) the distributive property
2
b) the value of i  1
c) when reducing a fraction every term is
affected by a common factor or a
negative sign. In this example “-4 ”
Example 7: Real and imaginary coefficients assigned to the variable with real and
imaginary numbers
3xi  6  5x  2i  4(2ix  3)  2x  7i
1. Remove any parenthesis.
2. Isolate the variables on the left
3xi  6  5x  2i  8ix  12  2x  7i
side and the real and imaginary
3xi  5x  8ix  2x  6  2i  12  7i
numbers on the right side of the
equation.
5ix  3x  6  5i
3. After simplifying the equation,
x(5i  3)  6  5i
remove the common factor “x”
(6  5i) (5i  3) 30i  18  25i 2  15i from the left side of the
x


equation resulting in a complex
2
(5i  3) (5i  3)
25i  9
number factor (-5i - 3).
4. When solving for one positive
30i  18  25(1)  15i 45i  43
x

“x” you must divide by the
25(1)  9
34
complex number factor and
45i  43  following the rule of no imaginary number in the denominator, the
SS 

 denominator and numerator on the right side must be multiplied by
34


(-5i + 3).
5. Simplify the resulting fraction. Some basic items to consider:
a) binomial expansion
2
b) the value of i  1
c) when reducing a fraction every term is affected by a common factor
or a negative sign. In this example “a negative sign”.
Compound Equations
By definition - an equation containing two different variables - one with real coefficients
assigned and the second having assigned imaginary coefficients.
Goal - to create two equations by isolating the real equation from the imaginary and solving
each equation for each particular variable following the procedures outlined for simple
equations. (Check procedure by referring to identified example)
Example 1:
3x  2i  6iy  5x  6  9iy
Imaginary Equation
Real Equation
3x  5x  6
3x  5x  6
2x  6
6
x
 3
2
Check:
The real equation formed
by equating the real
values from the left with
the real values on the
right side of the equal
sign. (Check example 1)
2i  6iy  9iy
6iy  9iy  2i
3iy  2i
2i 2
y

3i
3
3(3)  2i  6i 2 3  5(3)  6  9i  2 3
The imaginary equation
formed by equating the
imaginary values from
the left with the
imaginary values on the
right side of the equal
sign (Check example 3)
9  2i  4i  15  6  6i
9  6i  9  6i
Therefore, the complex number on the left side = the complex number on the right
Solution Set 3, 2 3
Example 2
Remember to remove
3(3ix  5)  2(2y  4i)  7ix  6i  7y  3 parenthesis
9ix 15  4y  8i  7ix  6i  7y  3
Imaginary Equation
Real Equation
9ix  8i  7ix  6i
9ix  7ix  8i  6i
2ix  2i
2i
x
 1
2i
Check
The imaginary
equation formed by
equating the imaginary
values from the left
with the imaginary
values on the right side
of the equal sign
(Check example 3)
15  4y  7y  3
4y  7y  15  3
3y  18
18
y
6
3
The real equation
formed by equating
the real values from
the left with the real
values on the right
side of the equal sign.
(Check example 1)
3(3i[1]  5)  2(2[6] 4i)  7i[1]  6i  7[6] 3
9i  15  24  8i  7i  6i  42  3
i  39  i  39
Therefore, the complex number on the left side = the complex number on the right
Solution Set {(-1, 6)}
System of Equations
By definition - an equation containing two different variables - each with real and imaginary
coefficients assigned.
Goal - a) to create two equations by isolating the real equation from the imaginary
following procedure of real values equal real values and imaginary values equal
imaginary values
b) to simplify each equation using procedures of simple equations
c) to solve the two equations as a system using the procedures of elimination or
substitution. Review procedures by referring to Solving Systems of Equations
5ix  3y  7  2iy  5x  4i
Example 1
Imaginary Equation
Real Equation
3y  7  5x
5x  3y  7
After each equation has been
formed, each is simplified and
put into the form of Ax + By = C
6ix  2iy  4i
6ix  2iy  4i
6x  2y  4
Solving the elimination - removing the x and then solving for one positive ‘y”
The system of equations
5x  3y  7  6(5x  3y  7)  30x  18y  42 Solving for “y”
Substitution
of 11 4
6x  2y  4  5(6x  2y  4)  30x  10y  20
6x - 2(11 4) = -4
8y = 22
for “y”
and
solving
for “x”
Check
4  6x - 4  2(11 4) = 4  -4
y = 22 8  11 4
24x  22  16
24x  6
x  6 24  1 4
5ix  3y  7  2iy  5x  4i
   
   
   
 
5i 1 4  3 11 4  7  2i 11 4  5 1 4  4i
4  5i 1 4  4  3 11 4  4  7  4  2i 11 4  4  5 1 4  4  4i
5i  33  28  22i  5  16i
5i  5  5i  5
 
Therefore, the complex number on the left side = the complex number on the right
1 , 11
Solution Set
 4 4
The next example will use the process of substitution to solve the system of equations
3x  4iy  7  5xi  5y  6i
Example 2
Imaginary Equation
Real Equation
3x  7  5y
3x  5y  7
After each equation has
been formed, each is
simplified and put into the
form of Ax + By = C
4yi  2xi  6i
5xi  4iy  6i
5x  4y  6
The system of equations
3x  5y  7  3x  5y  7  x 
5y  7
3
Solving for “x”
5x  4y  6
Substitution
of 5y  7
3
for “x” and
solving for
one positive
“y”
5y  7 
5
 4y  6

3
5y  7 
3  5
 3  4y  3  6

3
25y  35  12y  18
13y  53
y  5313
 
3x - 5 5313  7
 
Substitution of
53 for “y”
13
and solving
for one
positive “x”
13 3x -13  5 5313  13  7
39x - 265 = -91
39x = 174
x = 174 39  58 13
Check

  
   
 
 
 
3 5813  4i 5313  7  5 58 13 i  5 5313  6i
13  3 5813  13  4i 5313  13  7  13  5 5813 i  13 5 5313  13 6i
174  212i  91  290i  265  78i
 
265  212i  212i  265
Therefore, the complex number on the left side = the complex number on the right
Solution Set 58 , 53 

13
13

Quadratic Equations
The previous examples contained a visual “i” and it was easy to identify each equation as
one that contained imaginary or complex numbers. The quadratic equation with real
coefficients hides any reference to the existence of imaginary or complex numbers until one
is asked to solve (determine the roots, find the critical zeros, calculate the x-intercepts) the
equation. The procedure used most often is the application of the quadratic formula:
b  b 2  4ac
x
2a
and in particular, the discriminant b 2  4ac. If the value of the discriminant is less than zero
(a negative value) the nature of the roots can be interpreted as two unequal and imaginary
roots. A further interpretation is that a graph of this function would not have any xintercepts or critical zeros. One other note is that imaginary roots always occur in pairs
and are conjugates on one another.
Example 1:
Determine the roots of the quadratic equation
x 2  3x 11  0
b  b  4ac
x
; a 1, b  3, c  11
2a
Note:
1. Take time when writing out and substituting
values into the formula paying particular
attention to negative signs.
2
2. Use the concept of i  1 to remove the
negative values from under the radical sign
3. Remove all perfect from under the radical.
2
3  (3)2  4(1)(11)
x
2(1)
3  9  44 3  35 3  i 35
x


2
2
2
Once the roots have been determined one task remains and that is to check whether the
calculated roots are the correct values. Two distinct procedures can be used.
a) The first procedure requires the substitution of a root into the original equation,
simplification and a result of 0 = 0.
b) The second procedure utilizes two parts
i) the formulas -b/a and c/a which equals the sum and the product of the roots
respectively.
ii) the addition of and multiplication of the calculated roots
iii) comparison of the results from i) and ii)
Procedure One
To help simplify the equation, remove
2
3  i 35  3  i 35 
the fractions be multiplying each term
2
x  3x  11  0  

3

11

0
 

by a common denominator.Remember
 2
  2

to use the distributive property on the
2
9  3i 35  3i 35  i 35 9  3i 35
second fraction.

 11  0
4
2
9  6i 35  35 9  3i 35
6i 35  26 
9  3i 35 

 11  0  4  

4



  4 11  4  0
4
2

4


2

Note:
6i 35  26  18  6i 35  44  0  0  0
1. Adding conjugates results in
the radical term being
Procedure Two
eliminated.
b
3
c
11
2
2. When multiplying conjugates
x  3x  11  0  s1      3, p1    11
a
1
a 1
the middle term is always
3  i 35  3  i 35  6
eliminated.
s2  



3
 

3. A radical times itself always

2
 
2
 2
results in an answer of the
2
3  i 35 3  i 35  9  i 35 9  35 44
radicand.
p2  



 11




2

2

4
4
4
35  35  35
s1  s2, p1  p2
3  i 35 3  i 35 
Solution Set: 
,

2
2


 35  35  35
i 35  i 35  35i 2  35
i 35  i 35  35i 2  35
Example #2:
Determine the roots of the quadratic equation
b  b  4ac
x
; a  3, b  5, c  7
2a
- 3x 2  5x  7  0
2
5  (5)2  4(3)(7)
x
2(3)
x
Note:
Always pay attention to the substitution
of negative values for a, b and c
5  25  84 5  59 5  i 59


6
6
6
Check:
2
5  i 59  5  i 59 
3x 2  5x  7  0  3
  5
 7  0
 6   6 
25  5i 59  5i 59  i 2 59  5  i 59 
25  10i 59  59  5  i 59 
3

5

7

0

3
 


  5
 7  0

4
  6 

36
  2

30i 59  102 
25  5i 59 
75  30i 59  177 25  5i 59

 7  0  36  

36



 36  7  36 0
36
6

36


6

30i 59  102  150  30i 59  252  0  0  0
Be careful with sign changes when
simplifying this part of the equation
Check 2:
b
5 5
c 7 7
2
3x  5x  7  0  s1      , p1  

a
3 3
a 3 3
5  i 59  5  i 59  10 5
s2  

  
 
 6
  6
 6 3
s1  s2, p1  p2
5  i 59 5  i 59  25  i 2  59 25  59 84 7
p2  






 6

 6

36
36
36 3
Solution Set
5  i 59 5  i 59 
,


6
6


Determining a Quadratic Equation
The ability to calculate the roots of a quadratic equation also allows for the mathematical
procedure of determining a quadratic equation given the roots of an equation. Two
procedures can be used. The include:
a) writing the equation as a product of factors and then simplifying the equation
b) using the formula x 2  (r1  r2 )x  r1  r2  0
where r1 and r2 represent the given roots
The following examples will demonstrate the two procedures. Evaluate both procedures and
decide which one is best suits your mathematical skills.
Example 1: non fractional roots
Determine the quadratic equation that has a solution set of
3  i
5,3  i 5

Procedure 1 - writing as factors
x - 3  i 5x - 3  i 5  0  x  x3  i 5  x3  i 5  3  i 5 3  i 5 0
2
x  3x  ix 5  3x  ix 5  9  i  5  0  x  6x  9  5  0
2
2
2


x  6x  14  0
2
Procedure 2 - using the formula
 


x 2  (r1  r2 )x  r1 r2  0  x 2  3  i 5  3  i 5 x  3  i 5 3  i 5  0
x 2  6x  9  i 2  5  0  x 2  6x  9  5  0
x 2  6x  14  0
When the roots are not fractions the amount of work required does not appear to be that
much different but additional work is still required when using binomial expansion in the
first procedure as identified by the red boxes.
Example 2: fractional roots
Determine the quadratic equation that has a solution set of
2  i 7 2  i 7 
,


5
5


Procedure 1 - writing as factors
 2  i 7  2  i 7 
 2  i 7   2  i 7   2  i 7  2  i 7 
2
x
x

0

x

x
 
 


 x     0

5


5


5


5


5


5








2x  ix 7 2x  ix 7 4  i  7
2x  ix 7 2x  ix 7 4  7
x 


 0  x2 


0
5
5
25
5
5
25
4x 11
4x
11
x 2    0  25  x 2  25  25  25 0  25x 2  20x 11  0
5 25
5
25
Procedure 2 - using the formula
2
2
2  i 7 2  i 7  2  i 7 2  i 7 
x 2  (r1  r2 )x  r1 r2  0  x 2  

x  
0




 5
5   5  5 
4  4  i 2 7 
4 11

2
x 
x  

0

x

x 0
 5   25 

5 25
2
4
11
25  x 2  25 x  25   25 0  25x 2  20x 11  0
5
25
This example demonstrates that the first
procedure requires a lot more work
especially when the binomial expansion
involves fractions as shown by the red
boxes. The last part of both procedures
requires the same amount of calculation
Example 3: the roots written as a combined expression
Determine the quadratic equation that has a solution set of
Procedure 1 - writing as factors
4  i 11 


6


To determine the equation it
is necessary to identify the
given 2 roots
4 i 11 4  i 11 
,


6
6


 4  i 11  4  i 11 
 4  i 11   4  i 11   4  i 11  4  i 11 
2
x

x


0

x

x
 
 


 x 
  

  0

6


6


6


6


6


6








4x  ix 11 4x  ix 11 16  i 121
4x ix 11 4x  ix 11 16 121
x 


 0  x2 


0
6
6
36
6
6
36
8x 137
8x
137
x2  
 0  36 x 2  36   36 
 36  0  36x 2  48x 137  0
6 36
6
36
Procedure 2 - using the formula
2
2
4 i 11 4  i 7  4  i 11 4  i 11 
x 2  (r1  r2 )x  r1 r2  0  x 2  

x  
0




 6
6   6  6 
8 16 i 2 121
8 137

2
x 
x  

0

x

x
0
 6   36 

6
36
2
8
137
36  x 2  36 x  36 
 36  0  36x 2  48x 137  0
6
36
If not indicated, the choice of which
procedure (factors or formula) is left up
to you. Make your choice based on
mathematical skill and chance for best
success in determining the equation
A special application of imaginary numbers
A quick focus on factoring:
x 2  4  (x  2)(x  2)
9x 2  25  (3x  5)(3x  5)
Each of these examples represent a difference of squares
and the factors are a set conjugates formed by taking the
square root of each of the values in the question.
2
2
2
Until now questions of the form x  4, x  25, 9x  49 have not been factorable.
The existence of imaginary numbers permits the factoring of an equation because of this
which in term allows
basic fact
2
2
16

16i
 4i  4i
positive perfect squares to
2i  2i  4i  4
2
be re-written as a product
2
25

25i
 5i  5i
8i  8i  64i  64
as indicated in these
2
144  144i 2  11i 11i
11i 11i  121i  121 examples
2
2
To factor questions of the form x  y
Step 1: re-write the question so that appears as a difference of squares (this is accomplished
by using the above fact)
Include binomial multiplkicatiom