At the conclusion of this lesson you will be able to:

• Write complex numbers in the form of a + bi

• Find zeros of quadratic functions involving complex numbers

• Determine how many complex roots a polynomial may have before factoring it completely

• Write polynomial functions that have complex roots.

We have seen several times where a graph does not intersect the x-axis.

This means that there is no real number that makes a function equal to 0.

Ex: 4)( 2 xxf

To examine the zeros of a function like this we must extend our numbersystem once more to the set of Complex Numbers.

Before we can define complex numbers, we must introduce the number .i

1i From this we get 12 i

cannot be a real number because there is no real number that satisfiesi

012 x

Complex Numbers

The set of complex numbers consists of all expressions of the form

,biaz Where a and b are real numbers. This set is often denoted C.

The number a is called the real part of the complex number z, and thenumber b is called the imaginary part of z.

When b = 0, the complex number is a real number, therefore, all real numbersare complex numbers.

Two complex numbers are equal if and only if their real parts are equal and theirimaginary parts are equal.

a + bi = c + di if and only if a = c and b = d

Addition, Subtraction and Multiplication of Complex Numbers

Addition and subtraction is done just like other numbers: combine like terms.

Ex 1: Determine the sum and product of the complex numbers:

Multiplication is just like multiplying binomials: FOIL method.

iandi 7532

Sum: )75()32( ii

i

ii

47

7532

Difference: )75()32( ii

i

ii

103

7532

Product: )75)(32( ii 221151410 iii

)1(2110 i2110 i i31

Recall: quadratic formula:a

acbbx

242

We use the quadratic formula to find the zeros of a quadratic function.

When there are two real solutions. 042 acb

When there is one real solution.042 acb

When there are two complex solutions.042 acb

042 acb

How to recover the complex roots from the quadratic formula:

04 2 bac Multiplied both sides by -1

24 bac This is now a real number

)1)(4( 2bac acb 42 14 2bac ibac 24

Ex 2: Determine all the solutions to the equation 01342 xx

Solution: a = 1 b = -4 c = 13

2

)13)(1(444 2 x

252164 x

ix2

)1(364

ix2364

ix264

ix26

24

ix 32

ixandix 3232

Recall: turning a zero into a factor:

If x = 3, the factor would be…(x – 3).

Now, turn the complex roots into factors.

))32(())32(( ixandix

NOTE: The two complex solutions have the same real part and the imaginary parts only differ by sign. These are called Complex Conjugates.

))32(())32((1342 ixixxx

Complex Conjugates

The complex conjugate of the complex number

is the complex number

biaz

.biaz

For any complex number z,

abiabiazz 2)()(

22))(( babiabiazz

which are both real numbers.

Ex 3: Determine the complex conjugates of and and then find:

iz 34 iw 52

wzwzwzwz ,,,

Solution: iz 34 iw 52

iiiwz 26)52()34(

iiiiwz 2626)52()34(

iiiiiiwz 1423156208)52)(34( 2

iiiiiiiwz 14231423156208)52)(34( 2

Conjugate Results

For any pair of complex numbers, z and w, and for any integer n, we have

nnzzandwzwzwzwz ,

Complex roots come in conjugate pairs.

Ex: If x = 2 – 3i is a complex root of a polynomial, then so is x = 2 + 3i.

Ex 4: One solution to the equation is the

complex number . Determine all the solutions to this equation.

0522146 234 xxxx

ix 21

Solution: If x = 1 + 2i is a solution then so is x = 1 – 2i.

Now, turn them into factors and simplify.

(x – (1 + 2i)) = (x – 1 – 2i)(x – (1 – 2i)) = (x – 1 + 2i)

The product of these two new factors is the divisor (D(x)) for yourdivision algorithm.

(x – 1 – 2i)(x – 1 + 2i) = x2 – 2x + 5 = D(x)

Notice that when complex conjugates are multiplied that the i’scancel out.

Now, divide using the division algorithm.

1452214652

2

2342

xxxxxxxx

Factor Q(x)

We can see that we need to use the quadratic formula a = 1 b = -4 c = 1

322324

21242

4164

)1(2

)1)(1(4)4(4 2

x

x

x

x

x32

32

x

x

turn solutions into factors.

)32((

))32((

x

x

What if the directions asked to factor completely?

ixixxxxxxx 21213232522146 234

Ex 5: Factor completely. 2793)( 23 xxxxf

Solution: We need to find at least one zero on our own before we can do the division algorithm. (RATIONAL ZERO TEST)

9,3,11

9,3,1

1

27

offactors

offactors

027)3(9)3(3)3()3( 23 f

x = 3, the factor is: (x - 3) which is your D(x) for your divisionalgorithm.

927933

2

23

xxxxx

Now use the quad. formulato find other zeros.

iii

x 32

6

2

36

2

36

12

91400 2

)3)(3)(3(2793)( 23 ixixxxxxxf