A.6 Complex
Numbers & Solving
Equations
Get busy living, or get busy dying.
-Red
Complex Numbers
Complex numbers are numbers of the form
a  bi
where a is called the real part and b is the imaginary part






i  1
i 2  1
i 3  i
i 4 1
i5  i
i19  ?
Complex Numbers
To add complex numbers, add the real parts and add
the imaginary parts together (same for subtraction).
(a  bi)  (c  di)  (a  c)  (b  d)i
(a  bi)  (c  di)  (a  c)  (b  d)i
To multiply complex numbers, follow the usual rules
 for multiplying two binomials. DOUBLE DISTRIBUTION!!!


(a  bi)  (c  di)  a(c  di)  bi(c  di)
i 2  1
Complex Numbers
Given the following complex numbers, perform the
z  1 4i
indicated operations. w  5  3i
1) w  z
2) w  z
3) w  z


Complex Numbers
If z  a  bi is a complex number, then its conjugate,
denoted by z , is defined as
z  a  bi  a  bi


zz  a2  b2

Complex Numbers
Let’s find the conjugate of the following complex numbers
and then multiply by the conjugate.
1) z  3 4i
2) z  1 8i
3) z  2i
Complex Numbers
We can apply multiplying conjugates to quotients of complex
numbers. Let’s write the following in standard form!
2  3i
4  3i

Complex Numbers
Given the following complex numbers, perform the
z  5  2i
indicated operations. w  2  3i
1) w  z
2) w  z
3) w  z
w
4)
z


Solving Equations
Let’s recall solving the following equation.
x2  4  0
If we extend our number system to include complex numbers,
we can solvethe previous equation. How cool is that!
If N is a positive real number, we define the principal square
root of –N, denoted by N , as
N  Ni

x   4 
N In Complex Number System
Evaluate the square root of the following negative numbers
in the complex number system.
1)
3
2)
 18
Equations in Complex Number System
Solve the following equations in the complex number system.
2x 3  3x 2  6x  9  0
x 3  6x 2 12x  0
2x 6  2x 4  24x 2  0
10x 2  6x 1 0
Equations in Complex Number System
2x 3  3x 2  6x  9  0
x  3 2 ,  3i
x 3  6x 2 12x  0
x  0,  3  3i
2x 6  2x 4  24x 2  0

10x 2  6x 1 0

x  0, 0,  3,  2i
3  i
x
10

Now that we have extended our number system to include
complex numbers, we should get as many solutions as the

degree of the polynomial we aretrying to solve!
Character of Solutions
For the equation ax 2  bx  c  0
b2  4ac  0 the equation has two unequal real solutions
2 
b  4ac  0 the equation has a repeated real solution
the equation has two complex solutions that
b  4ac  0 are not real and are conjugates of each other
2
Now look at your worksheet for determining the character of
the equations!
Character of Solutions
Ex. x  6 x  13  0
2
b2  4ac  0 the equation has two unequal real solutions
b2  4ac  0 the equation has a repeated real solution
b2  4ac  0
the equation has two complex solutions that
are not real and are conjugates of each other
Homework: p.1008
#9 – 45 EOO,
53 – 69 EOO & 71
A.6 Complex
Numbers & Solving
Equations
Get busy living, or get busy dying.
-Red