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COMPLEX Z R O S

COMPLEX Z R O S. Complex zeros or roots of a polynomial could result from one of two types of factors: Type 1 Type 2 Notice that with either type, the

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Page 1: COMPLEX Z R O S. Complex zeros or roots of a polynomial could result from one of two types of factors: Type 1 Type 2 Notice that with either type, the

COMPLEXZ

ROS

Page 2: COMPLEX Z R O S. Complex zeros or roots of a polynomial could result from one of two types of factors: Type 1 Type 2 Notice that with either type, the

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

042 x 42 x ix 2Type 1

Type 2 0522 xx

ix 21

2

162

12

51422 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.

Page 3: COMPLEX Z R O S. Complex zeros or roots of a polynomial could result from one of two types of factors: Type 1 Type 2 Notice that with either type, the

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

ixixx 31312 Multiply the last two factors together. All i terms should disappear when simplified.

22 9333132 iixiixxixxx -1 1022 2 xxx Now multiply the x – 2 through

20144 23 xxxHere is a 3rd degree polynomial with roots 2, 1 - 3i and 1 + 3i

If x = the root then x - the root is the factor form. ixixx 31312

Page 4: COMPLEX Z R O S. Complex zeros or roots of a polynomial could result from one of two types of factors: Type 1 Type 2 Notice that with either type, the

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

20144 23 xxxxf 1, 2, 4, 5, 10, 20

1

Possible rational roots

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

Let’s try 2(synthetic division)

2 -4 202 1 -4 14 -20

1 -2 10 0Now put variables back in and factor or use quadratic formula

01022 xx

By quad formula: ix 31 So there was one positive and two imaginary roots

Page 5: COMPLEX Z R O S. Complex zeros or roots of a polynomial could result from one of two types of factors: Type 1 Type 2 Notice that with either type, the

Use the given root to find the remaining roots of the function

ixxxxxf 3root ;18452553 234 Since 3i is a root we also know that its conjugate -3i is also a root. Let’s use synthetic division and reduce our polynomial by these roots then.

9i -27+15i -45-6i 18 3i 3 5 25 45 -18

3 5+9i -2+15i -6i 0

3 5 -2 0

-3i -9i -15i 6i 0

0253 2 xxPut variables in here & set to 0 and factor or formula this to get remaining roots. 0213 xx

So the roots of this function are 3i, -3i, 1/3, and -2

Page 6: COMPLEX Z R O S. Complex zeros or roots of a polynomial could result from one of two types of factors: Type 1 Type 2 Notice that with either type, the

Acknowledgement

I wish to thank Shawna Haider from Salt Lake Community College, Utah USA for her hard work in creating this PowerPoint.

www.slcc.edu

Shawna has kindly given permission for this resource to be downloaded from www.mathxtc.com and for it to be modified to suit the Western Australian Mathematics Curriculum.

Stephen CorcoranHead of MathematicsSt Stephen’s School – Carramarwww.ststephens.wa.edu.au