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Imaginary and Complex Numbers
0 The imaginary number i is defined as 1
0 i2 = -1
0 An expression in the form a + bi is called a complex
number and has:
0 Real part “a”
0 Imaginary part “bi”
0 When adding/subtracting complex numbers, you can only
combine the like parts – real with real and imaginary with
imaginary
0 When multiplying binomials containing imaginary parts,
using FOIL
0 A fraction containing an imaginary part in the denominator
must be rationalized
Imaginary and Complex Numbers
0 Simplify 4
0 We can rewrite this as 1  4
0 This will simplify to 2i
0 Simplify 18
0 We can rewrite this as 1  9  2
0 This simplifies as 3i 2
0 Simplify (3 + 2i) + (5 – 7i)
0 Combine like terms: 8 – 5i
Imaginary and Complex Numbers
0 Simplify (2 + 3i)(3 – 2i)
0 6 – 4i + 9i – 6i2
0 6 + 5i – 6i2
 Remember that i2 = -1
0 6 + 5i – 6(-1)
0 6 + 5i + 6
0 12 + 5i
Imaginary and Complex Numbers
0 Simplify 3  2i
0 3  2i i 5i

5i
i
0 3i  2i2
 Multiply by a form of 1 to get rid of the
imaginary number in the denominator
5i2
0 3i  2  1
5  1
2  3i
0 2  3i
or 
5
5
Imaginary and Complex Numbers
0 Simplify 5  2i
3  4i
0 Because the denominator is a binomial, we cannot
0
0
0
0
simply multiply by i/i to rationalize the denominator
We need to multiply by the conjugate of the
denominator
The conjugate is the binomial that will create a
difference of squares when multiplied by the
denominator
To find the conjugate, simply change the sign of the
second term of the denominator
For this problem, the conjugate would be 3 – 4i
Imaginary and Complex Numbers
0 Simplify 5  2i
 Multiply by a form of 1 using the conjugate to get
rid of the imaginary number in the denominator
0 5  2i 3 34i 4i

3  4i 3  4i
0 15  20i  6i  8i2  Because the conjugate creates a difference of
9  12i  12i  16i2 squares, the middle terms cancel out
0
15  14i  8  1
9  16  1
15  14i  8
0 9  16
0
23  14i
25
Solving Equations
0 Solve: -3x2 = 75
0 x2 = -25
0 x2   25
0 x = ±5i
0 Solve: 2x2 – 4x + 9 = 0
0
b  b2  4ac
x
2a
0 x
0
  4  
x
 4   4 2  9 
22
4  16  72
4
x
4  56
4
x
4  1  4  14
4
2
4  2i 14
x
4
x
2  i 14
2
Quadratics and Complex Numbers
0 Determine if 2x2 – 3x + 5 = 0 has real or imaginary
roots.
0 We can use the discriminate (b2 – 4ac) from the
quadratic formula to determine if the roots are real or
imaginary without having to actually solve the equation.
0 If the discriminate is POSITIVE, the roots will be REAL
numbers
0 If the discriminate is NEGATIVE, the roots will be
IMAGINARY/COMPLEX numbers
Quadratics and Complex Numbers
0 Determine if 2x2 – 3x + 5 = 0 has real or imaginary
roots.
0 The discriminate is b2 – 4ac
0 (-3)2 – 4(2)(5)
0 9 – 40
0 -31
0 As the discriminate is negative, this quadratic will have
IMAGINARY/COMPLEX roots
Quadratic Functions
0 Look at the function below and discuss its zeros.
y














x





















We know that all quadratic functions
must have two zeros, and that the zeros
are the points where y = 0. Generally,
this is where the graph crosses the
x-axis. The given graph does not cross
the x-axis at all, but must still have two
zeros. This means that the zeros of the
given graph must be imaginary.



**NOTE: IMAGINARY ZEROS AND ROOTS MUST
ALWAYS OCCUR IN PAIRS.**