Transcript Imaginary Numbers and The Fundamental Theorem of Agebra

Imaginary Numbers and The
Fundamental Theorem of Agebra
The Imaginary Unit
The imaginary unit, or i, is the number
where:
i² = −1.
Imaginary Numbers
• A real number multiplied by i, where i is a
number so that i² = -1 .
About Numbers…
• Real Numbers: Can be plotted on a
number line
• Imaginary numbers: A real number times
the imaginary number i, where i is a
number so that i² = -1 .
• Complex Numbers: All real and imaginary
numbers
Complex numbers
• Complex numbers, written in standard
form, are a + bi where i is an imaginary
number and a and b are real numbers.
• If b = 0, then the number is just a, and is a
real number.
• If b != 0, then the number is an imaginary
number.
• If a = 0, then it is a pure imaginary
number.
Working with Imaginary Numbers:
Addition and Subtraction
• Sum:
(a + bi) + (c + di) = (a + c) + (b+d)i
• Difference
(a + bi) – (c + di) = (a – c) + (b – d)i
Multiplying Complex Numbers
• Multiplying complex numbers works the
same as for real numbers
Exponents of i
Repeats what happens in
The blue section here --->
Exponents of i
To make things easier, we an think of i as a
left turn.
If we don’t turn left at all (i0), we face 1.
If we turn left once, we face i.
If we turn left twice, we are in the opposite
direction of where we started: -1.
If we turn left 3 times, we are in the opposite
direction of one turn: -i
Fundamental Theorem of
Algebra
y  x  3x  16 x  12
3
2
The largest power of x tells you the
total number of zeroes (both real and
imaginary). We call this the number
of complex zeroes.
For this problem, there are 3 complex
zeroes.
Finding the zeroes
y  x  3x  16 x  12
3
2
• Before we start finding the zeroes, we
need to learn as much as we can about
the possible number of zeroes. This can
save us a lot of time – if we know there are
no negative zeroes possible, then we don’t
have to try to find them.
Finding the zeroes
y  x  3x  16 x  12
3
2
• The # of possible positive zeroes is the
number of changes in the sign of f(x).
• For this function, the signs are:
+--There is only one change in sign, so there
is, at most, 1 positive zero.
Finding the zeroes
y  x  3x  16 x  12
3
2
• To find the number of possible negative
zeroes, we need to look at the number of
changes in the sign of f(-x). That means
we need to change the signs of the odd
powers, then look at the changes in sign.
Finding the zeroes
f ( x)  x  3x  16 x  12
3
2
f ( x)   x  3x  16 x  12
3
2
The signs for f(-x) are:
- -+The sign changes twice, so there are (at
most) 2 negative zeroes.
Finding the possible zeroes
• To find which zeroes our possible zeroes
could be, we can divide the factors of our
constant by the factors of our lead
coefficient:
3
2
y  x  3x  16 x  12
Constant Factors
Lead factors
+ 1, 2, 3, 4, 6, 12
+1
After we have the possible
zeroes…
1. We need to test each one.
2. Once we find one, we can use synthetic
division to get rid of that zero.
3. Repeat until we have a quadratic
(ax² + bx + c)
4. Factor to get the last few zeroes.
Finding the zeroes
y  x  3x  16 x  12
3
2
Possible Zeroes: + 1, 2, 3, 4, 6, 12. At most, 1 zero is positive. At most,
one zero is negative.
Test: x = 1
x² -5x -6
We have a quadratic left, so we can factor to find the last two zeroes:
(x-6) (x+1) = 0
x = 6, x = -1, or x = -2
Check your answer:
Just like our possible zeroes predicted, we have two negative zeroes
and one positive zero.