complex numbers 1.5 true or false: all numbers are complex numbers
TRANSCRIPT
Complex numbers 1.5
True or false:
All numbers are Complex numbers
POD
Consider our final equation from the previous lesson. We determined that x = -2 and x = 1 were solutions.
How many solutions will there be, counting multiplicities and complex solutions?
Are -2 and 1 the only solutions, or are there other solutions as well?
How could we find out?
087 36 xx
POD
Remember that substitution? Let’s factor with it.
Ooh, a difference of cubes and a sum of cubes. How do you factor those?
0)1)(8(
0)1)(8(
087
087
33
2
36
xx
mm
mm
xx
PODLet’s do a complete factorization. Finding all the
factors helps us find all the solutions.
We can see our two solutions easily. Do they have a multiplicity greater than one? (CAS does this easily.)
How do we find the other solutions?
0)1)(1)(42)(2(
0)1)(8(
087
22
33
36
xxxxxx
xx
xx
PODTurns out we have some imaginary solutions.
Use the quadratic formula to find them.
2
31
2
31
2
)1)(1(411
1
312
322
2
122
2
)4)(1(442
42
0)1)(1)(42)(2(
2
2
22
ix
xx
ii
x
xx
xxxxxx
PODAll six solutions:
2
31
31
1
2
0)1)(1)(42)(2( 22
ix
ix
x
x
xxxxxx
A brief review of i
What is i?
What would i2 equal?
What about i3 or i8?
A brief review of i
What is i?
What would i2 equal? -1
What about i3 or i8?
1
1
18
3
i
ii
A brief review of their form
a + bi(real component) (imaginary
component)
What do you have when a = 0?
What about when b = 0?
Equivalent complex numbers
a + bi = c + di only if a = c and b = d
Solve for x and y:(2x-4) + 9i = 8 + 3yi
Equivalent complex numbers
a + bi = c + di only if a = c and b = d
Solve for x and y:(2x-4) + 9i = 8 + 3yi2x-4 = 8 9 = 3yx = 6 y = 3
Adding complex numbers
(a + bi) + (c + di) = (a + c) + (b + d)i
Add: (3 + 4i) + (2 - 5i)
Subtract: (3 + 4i) - (2 - 5i)
Try this on calculators.
Adding complex numbers
(a + bi) + (c + di) = (a + c) + (b + d)i
Add: (3 + 4i) + (2 - 5i) = 5 - i
Subtract: (3 + 4i) - (2 - 5i) = 1 + 9i
Try this on calculators.
This is how it looks on the TI-84.
(a + bi) + (c + di) = (a + c) + (b + d)i
Add: (3 + 4i) + (2 - 5i)
Subtract: (3 + 4i) - (2 - 5i)
Multiplying complex numbers
(a + bi)(c + di) = (ac - bd) + (ad + bc)i
Multiply: (3 + 4i)(2+5i)
Multiply: (3 - 4i)(2+5i)
Try this on calculators.
Multiplying complex numbers
(a + bi)(c + di) = (ac - bd) + (ad + bc)i
Multiply: (3 + 4i)(2+5i) = -14 + 23i
Multiply: (3 - 4i)(2+5i) = 26 + 7i
Try this on calculators.
Complex conjugates
What is the complex conjugate of a + bi?
What is the product of a + bi and its complex conjugate?
Complex conjugates
What is the complex conjugate of a + bi?
What is the product of a + bi and its complex conjugate?
That means the factorization for is
22 ba
22 ba
))(( biabia
Operations with complex numbers
Express in a + bi form:4(2 + 5i) - (3 - 4i)
(4 - 3i)(2 + i) (3 - 2i)2
i(3 + 2i)2
i51
Operations with complex numbers
Express in a + bi form:4(2 + 5i) - (3 - 4i) = 5 +24i
(4 - 3i)(2 + i) = 11 – 2i (3 - 2i)2 = 5 - 12i i(3 + 2i)2 = -12 + 5i i51 = i3 = -i
Rational expressions with complex numbers
Simplify
Hint: What is the complex conjugate of the denominator?
And try this on calculators. How do you get rational coefficients?
7 i3 5i
Rational expressions with complex numbers
Simplify
i
iii
iii
i
i
i
i
17
16
17
1317
1613
34
3226259
533521
)53(
)53(
)53(
)7(
2
2
Complex numbers as radical expressions
Multiply
Hint: Rewrite using i.
5 9 1 4
Complex numbers as radical expressions
Multiply.
Without using i, we’d have three different radicals, and wind up with a different real number component.
i
iii
ii
131
63105
2135
4195
2