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