REVIEW OF COMPLEX NUMBERS
1  i and
The Imaginary Number i :
i 2  1
Complex numbers are always written in the form a + bi
part)
(a is the real part and b is the imaginary
Examples: Write as complex numbers.
a)
25 =
25 
1 25 
1  25  5i
b)
17 =
17 
117 
1  17  i 17
c)  9 =  9  


1 9  


1  9    3 i    3 i
d) (2  3i )  ( 4  6i) = 2  3i   4  6i  2  4  3i  6i   2  9i
e) (6  10i )  (9  3i) = 6  10i  9  3i  6  9  10i  3i  15  13i
f) (6  4i )  (2  6i) = (6  4i)  (2  6i)  6  4i  2  6i  8  2i
g) (2  i)  (3  10i) = (2  i)  (3  10i)  2  i  3  10i   1  9i
h) (8  3i )(1  2i ) = 8  16i  3i  6i 2  8  19i  6(1)  8  19i  6  2  19i
i) 3i (5  6i) = 15i  18i 2  15i  18(1)  15i  18  18  15i
The conjugate of a + bi
is a – bi .
Examples: Find the conjugate of the following complex numbers.
a) 2 + 6i
 2 – 6i
b) –3 – 4i
 –3 + 4i
c) 15i
 –15i
To divide two complex numbers, multiply the numerator and denominator by the conjugate of
the denominator.
Examples: Divide. Be sure to write each answer in standard form.
a)
5
5  2i 
5(2  i )


 
2i 2i  2i 
(2  i )(2  i )
10  5i

4 2i 2i  i 2
10  5i

4  (1)
10  5i

5
10
5i


5
5
 2i
b)
1  4i 1  4i  5  3i 
(1  4i )(5  3i )


 
5  3i 5  3i  5  3i 
(5  3i )(5  3i )
5  3i  20i  12i 2
25 15i 15i  9i 2
5  23i  12(1)

25  9(1)
7  23 i

34
7
23 i


34
34

Powers of i :
i=
i
i5 =
i
i2 =
–1
i6 =
–1
i3 =
–i
i7 =
–i
i4 =
1
i8 =
1
So, the pattern for the powers of i is:
{ i, –1, –i, 1 }
Examples: Simplify
a) i12 = 1
(count through the pattern until you get to 12)
b) i67 = –i
(count through the pattern until you get to 67)
c) 4 + i3 = 4 + (–i) = 4 – i
d) 4i4 – 2i2 + i = 4(1) – 2(–1) + i = 4 + 2 + i = 6 + i
When solving the quadratic equation, ax 2  bx  c  0 , if b2 – 4ac < 0, then there will be 0 real
solutions (because the 2 answers are complex)
Examples: Solve
a)
x2  2 x  5  0
a = 1, b = –2, c = 5
2  (2)2  4 1 5
2  4  20
2  16
2  4i
x



 2  2i
2(1)
2
2
2
b) 6 x 2  4 x  1  0
a = 6, b = 4, c = 1
x
4  (4)2  4  6 1
4  16  24
4  8
4  2 2 i
4 2 2 i
1
2i







2(6)
12
12
12
12
12
3
6
Notice that complex solutions ALWAYS come in complex conjugate pairs!!
Example:
If 5i is a solution to a quadratic equation, then so is – 5i
If –4 – 6i is a solution to a quadratic equation, then so is –4 + 6i