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Complex Numbers
Consider the quadratic equation x2 + 1 = 0.
Solving for x , gives x2 = – 1
x2   1
x  1
We make the following definition:
i  1
Complex Numbers
i  1
2
Note that squaring both sides yields: i  1
therefore i 3  i 2 * i1  1* i  i
and i 4  i 2 * i 2  (1) * (1)  1
so
and
i  i * i  1* i  i
5
4
i  i * i  1* i  1
6
4
And so on…
2
2
Real numbers and imaginary numbers are
subsets of the set of complex numbers.
Real Numbers
Imaginary
Numbers
Complex Numbers
Definition of a Complex Number
If a and b are real numbers, the number a + bi is a
complex number, and it is said to be written in
standard form.
If b = 0, the number a + bi = a is a real number.
If a = 0, the number a + bi is called an imaginary
number.
Write the complex number in standard form
1   8  1  i 8  1  i 4  2  1  2i 2
Addition and Subtraction of Complex
Numbers
If a + bi and c +di are two complex numbers written
in standard form, their sum and difference are
defined as follows.
Sum: ( a  bi )  ( c  di )  ( a  c )  ( b  d
)i
Difference:( a  bi )  ( c  di )  ( a  c )  ( b  d
)i
Perform the subtraction and write the answer
in standard form.
( 3 + 2i ) – ( 6 + 13i )
3 + 2i – 6 – 13i
–3 – 11i
8   18  4  3i 2 
8  i 9  2  4  3i 2 
8  3i 2  4  3i 2
4
Multiplying Complex Numbers
Multiplying complex numbers is similar to
multiplying polynomials and combining like terms.
Perform the operation and write the result in
standard form.( 6 – 2i )( 2 – 3i )
F
O
I
L
12 – 18i – 4i + 6i2
12 – 22i + 6 ( -1 )
6 – 22i
Consider ( 3 + 2i )( 3 – 2i )
9 – 6i + 6i – 4i2
9 – 4( -1 )
9+4
13
This is a real number. The product of two
complex numbers can be a real number.
This concept can be used to divide complex numbers.
Complex Conjugates and Division
Complex conjugates-a pair of complex numbers of
the form a + bi and a – bi where a and b are
real numbers.
( a + bi )( a – bi )
a 2 – abi + abi – b 2 i 2
a 2 – b 2( -1 )
a2+b2
The product of a complex conjugate pair is a
positive real number.
To find the quotient of two complex numbers
multiply the numerator and denominator
by the conjugate of the denominator.
a  bi 
c  di 

a  bi  c  di 


c  di  c  di 
ac  adi  bci  bdi

2
2
c d
2
ac  bd  bc  ad i

2
2
c d
Perform the operation and write the result in
standard form.
6  7i 
1  2i 

6  7i  1  2i 


1  2i  1  2i 
6  14  5i
6  12i  7i  14i


2
2
1 4
1 2
2
20  5i

5
20 5i


5
5
 4 i
Perform the operation and write the result
in standard form.
1 i
3

1 i i
3 4  i 


 

i
4i
i
i 4  i 4  i 
i  i 12  3i   1  i  12  3i
 2  2 2
1
16  1
i
4 1
12
3
12 3
 1 i   i  1  i  i
17
17
17 17
2
17  12 17  3


i
17
17
5 14

 i
17 17
Expressing Complex Numbers
in Polar Form
Now, any Complex Number can be expressed as:
X+Yi
That number can be plotted as on ordered pair in
rectangular form like so…
6
4
2
-5
5
-2
-4
-6
Expressing Complex Numbers
in Polar Form
Remember these relationships between polar and
y
2
2
2
tan


rectangular form:
x y r
y  r sin 
x
x  r cos
So any complex number, X + Yi, can be written in
polar form: X  Yi  r cos  r sin i
r cos  r sin i  r (cos   i sin  )
Here is the shorthand way of writing polar form:
rcis
Expressing Complex Numbers
in Polar Form
Rewrite the following complex number in polar form:
4 - 2i
Rewrite the following complex number in
rectangular form: 7cis 30 0
Expressing Complex Numbers
in Polar Form
Express the following complex number in


rectangular form:
2 (cos
3
 i sin
3
)
Expressing Complex Numbers
in Polar Form
Express the following complex number in
polar form: 5i
Products and Quotients of
Complex Numbers in Polar Form
The product of two complex numbers,
r1 (cos1  i sin 1 ) and r2 (cos 2  i sin  2 )
Can be obtained by using the following formula:
r1 (cos1  i sin 1 ) * r2 (cos 2  i sin  2 )
 r1 * r2[cos(1   2 )  i sin( 1   2 )]
Products and Quotients of
Complex Numbers in Polar Form
The quotient of two complex numbers,
r1 (cos1  i sin 1 ) and r2 (cos 2  i sin  2 )
Can be obtained by using the following formula:
r1 (cos1  i sin 1 ) / r2 (cos 2  i sin  2 )
 r1 / r2[cos(1  2 )  i sin( 1  2 )]
Products and Quotients of
Complex Numbers in Polar Form
Find the product of 5cis30 and –2cis120
Next, write that product in rectangular form
Products and Quotients of
Complex Numbers in Polar Form
Find the quotient of 36cis300 divided by
4cis120
Next, write that quotient in rectangular form
Products and Quotients of
Complex Numbers in Polar Form
Find the result of (5(cos 120  i sin 120))
Leave your answer in polar form.
4
Based on how you answered this problem,
what generalization can we make about
raising a complex number in polar form to
a given power?
De Moivre’s Theorem
De Moivre's Theorem is the theorem which
shows us how to take complex numbers to any
power easily.
De Moivre's Theorem – Let r(cos F+isin F) be a
complex number and n be any real number. Then
[r(cos F+isin F]n = rn(cos nF+isin nF)
What is this saying?
The resulting r value will be r to the nth power and the
resulting angle will be n times the original angle.
De Moivre’s Theorem
Try a sample problem: What is [3(cos 45+isin45)]5?
To do this take 3 to the 5th power, then multiply 45 times 5
and plug back into trigonometric form.
35 = 243 and 45 * 5 =225 so the result is 243(cos 225+isin 225)
Remember to save space you can write it in compact form.
243(cos 225+isin 225)=243cis 225
De Moivre’s Theorem
Find the result of:
(1  i)
4
Because of the power involved, it would easier to change this
complex number into polar form and then use De Moivre’s Theorem.
De Moivre’s Theorem
De Moivre's Theorem also works not only for
integer values of powers, but also rational values
(so we can determine roots of complex numbers).
1
p
( x  yi )  (rcis  )
1
p
1

 r cis ( * )  r cis ( )
p
p
1
p
1
p
De Moivre’s Theorem
Simplify the following:
(4  4 3i)
1
3
De Moivre’s Theorem
Every complex number has ‘p’ distinct ‘pth’ complex
roots (2 square roots, 3 cube roots, etc.)
To find the p distinct pth roots of a complex number,
we use the following form of De Moivre’s Theorem
1
p
1
p
  360n
( x  yi )  r cis (
p
)
…where ‘n’ is all integer values between 0 and p-1.
Why the 360? Well, if we were to graph the complex
roots on a polar graph, we would see that the p roots
would be evenly spaced about 360 degrees (360/p would
tell us how far apart the roots would be).
De Moivre’s Theorem
Find the 4 distinct 4th roots of -3 - 3i
De Moivre’s Theorem
Solve the following equation for all complex
number solutions (roots): x 3  27  0