Pc Complex Zeros

Z COMPLEX R O S

Complex zeros of a polynomial could result from one of two types of factors:

x +4=0 2

Type 1

x = ± 2i

x2 + 2x + 5 = 0

Type 2

x=

x = −4 2

−2±

( 2) 2 − 4(1)( 5) 2(1)

− 2 ± − 16 = = −1 ± 2i 2

Notice that with either type, the complex zeros come in conjugate pairs. There are 2 complex solutions that have the same real term and opposite signs on imaginary term. This will always be the case. Complex zeros come in conjugate pairs.

So if asked to find a polynomial that has zeros, 2 and 1 – 3i, you would know another zero would be 1 + 3i. Let’s find such a polynomial by putting the zeros in factor form and multiplying them together. If x = the zero then x - the zero is the factor form.

( x − 2) ( x − (1 − 3i ) ) ( x − (1 + 3i ) )

( x − 2) ( x − 1 + 3i ) ( x − 1 − 3i ) ( x − 2) ( x

Multiply the last two factors together. All i terms should disappear when simplified.

− x − 3xi − x + 1 + 3i + 3 xi − 3i − 9i

2

(

= ( x − 2 ) x − 2 x + 10 2

)

2

)

-1

Now multiply the x – 2 through

= x 3 − 4 x 2 + 14 x − 20 Here is a 3rd degree polynomial with zeros 2, 1 - 3i and 1 + 3i

Let’s take this polynomial and pretend we didn’t know the zeros and work the other direction so we can see the relationship to everything we’ve learned in this chapter.

f ( x ) = x − 4 x + 14 x − 20 3

± Possible

2

1, 2, 4, 5, 10, 20

rational zeros

1

By Descartes Rule there are 3 or 1 positive real zeros (3 sign changes in f(x)) and no negative real zeros (no sign changes in f(-x)).

Let’s try 2 (of course from our previous work we know that will work!)

2

1 1

-4 2

14 -4

-20 20

-2

10

0

x 2 − 2 x + 10 = 0

By quad formula:

x = 1± 3i

Now put variables back in and factor or use quadratic formula

So there was one positive and two imaginary zeros

Use the given zero to find the remaining zeros of the function

f ( x ) = 3 x + 5 x + 25 x + 45 x − 18; zero = 3i 4

3

2

Since 3i is a zero we also know that its conjugate -3i is also a zero. Let’s use synthetic division and reduce our polynomial by these zeros then.

3i -3i

3

5 25 45 9i -27+15i -45-6i 3 5+9i -2+15i -6i -9i -15i 6i

-18 18 0 0

So the zeros3of this are 30i, -3i, 1/3, and -2 5 function -2 Put variables in here & set to 0 2 3x + 5 x − 2 = 0 and factor or formula this to get 3x − 1 x + 2 = 0 remaining zeros.

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