```A.6a Complex
Numbers
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  i
3





i4  1
i5  i
i19  ?
Complex Numbers
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
indicated operations. w  5  3i
z  1 4i
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
indicated operations. w  2  3i
z  5  2i
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)
36
2)
8
3)
32
Equations in Complex Number System
Solve the following equation in the complex number system.
2x 3  3x 2  6x  9  0
Equations in Complex Number System
Solve the following equation in the complex number system.
x 3  6x 2  12x  0
Equations in Complex Number System
Solve the following equation in the complex number system.
2x 6  2x 4  24 x 2  0
Equations in Complex Number System
Solve the following equation in the complex number system.
10x 2  6x  1  0
Equations in Complex Number System
2x 3  3x 2  6x  9  0
x  3 ,  3i
2
x 3  6x 2  12x  0
x  0,  3 3i
2x 6  2x 4  24 x 2  0
x  0, 0,  3,  2i
3 1
x
 i
10 10
10x 2  6x  1  0



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
b 2  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
Homework: p.1008
#9 – 31 odd,
39 – 45 odd, 24, 28
A.6a Complex
Numbers
Get busy living, or get busy dying.
-Red
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