Transcript Complex Numbers 1

Complex
Numbers 1
a + bi
We can add, subtract, multiply or divide complex numbers. After
performing these operations if we’ve simplified everything correctly we
should always again get a complex number (although the real or
imaginary parts may be zero). Below is an example of each to refresh
your memory.
ADDING
SUBTRACTING
MULTIPLYING
(3 – 2i) + (5 – 4i) = 8 – 6i
Combine real parts and
combine imaginary parts
3 – 2i - 5 + 4i = -2 +2i
Be sure to distribute the
negative through before
combining real parts and
imaginary parts
(3 – 2i) (5 – 4i)
FOIL and then combine like
terms. Remember i 2 = -1
(3 – 2i) - (5 – 4i)
= 15 – 12i – 10i+8i2
=15 – 22i +8(-1) = 7 – 22i
Notice when I’m done simplifying
that I only have two terms, a real
term and an imaginary one. If I
have more than that, I need to
simplify more.
DIVIDING
FOIL
2 1

15

12
i

10
i

8

3  2i 5  4i  15  12i  10i  8i


2 1


20
i

20
i

16

5  4i 5  4i 25
25  20i  20i  16i
Combine like terms
i 2  1
Recall that to divide complex numbers, you multiply
the top and bottom of the fraction by the conjugate of
the bottom.
23  2i 23 2


 i
41
41 41
We’ll put the 41 under
each term so we can
see the real part and
the imaginary part
This means the same
complex number, but
with opposite sign on
the imaginary term
Let’s solve a couple of equations that have complex
solutions to refresh our memories of how it works.
x  25  0
x    25
2
-25
2
-25
x   25 i  5 i
x  6 x  13  0
Use the
quadratic
formula
2
x
  6 
 6
21
2
The negative
under the square
root becomes i
 b  b 2  4ac
x
2a
 4113
6   16 6  16 i


2
2
Square root and
don’t forget the 
6  36  52

2
64i
 3 2i

2
Powers of i
ii
i  1
2
i  i i  1(i)  i
3
2
i  i i   1 1  1
4
2 2
i  i i  1i   i
6
4 2
i  i i  1 1  1
5
4
i  i i  1 i   i
7
4 3
i  i i  11  1
8
4 4
We could continue but notice
that they repeat every group
of 4.
For every i 4 it will = 1
To simplify higher powers
of i then, we'll group all the
i 4ths and see what is left.
  i  1 i  i
i  i
33
4 8
8
4 will go into 33 8 times with 1 left.
 
i  i
83
i  1 i  i
4 20 3
20 3
4 will go into 83 20 times with 3 left.
This "discriminates" or tells us what type of solutions we'll have.
ax  bx  c  0
2
 b  b  4ac
x
2a
2
If we have a quadratic equation and are considering solutions
from the complex number system, using the quadratic formula,
one of three things can happen.
1. The "stuff" under the square root can be positive and we'd get
two unequal real solutions if b 2  4ac  0
2. The "stuff" under the square root can be zero and we'd get one
solution (called a repeated or double root because it would factor
2
if
b
 4us
acthe
 0same solution).
into two equal factors, each giving
3. The "stuff" under the square root can be negative and we'd get
two complex solutions that are conjugates of
if each
b 2 other.
4ac  0
The "stuff" under the square root is called the discriminant.
The Discriminant
  b 2  4ac
Acknowledgement
I wish to thank Shawna Haider from Salt Lake Community College, Utah
USA for her hard work in creating this PowerPoint.
www.slcc.edu
Shawna has kindly given permission for this resource to be downloaded
from www.mathxtc.com and for it to be modified to suit the Western
Australian Mathematics Curriculum.
Stephen Corcoran
Head of Mathematics
St Stephen’s School – Carramar
www.ststephens.wa.edu.au