Z
COMPLEX
R
O
S
Complex zeros or roots 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   16

 1  2i
21
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 root would be 1 + 3i.
Let’s find such a polynomial by putting the roots in
factor form and multiplying them together.
If x = the root then
x - the root is the factor form.
x  2x  1  3i x  1  3i 
x  2x 1  3i x 1  3i 
x  2 x
2
Multiply the last two factors
together. All i terms should
disappear when simplified.
 x  3xi  x  1  3i  3xi  3i  9i

 x  2 x  2 x  10
2

2

-1
Now multiply the x – 2 through
 x  4 x  14 x  20
3
2
Here is a 3rd degree polynomial with roots 2, 1 - 3i and 1 + 3i
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.
f x   x  4 x  14 x  20
3

Possible
2
1, 2, 4, 5, 10, 20
rational roots
1
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
2
1
(synthetic
division)
1
-4
2
-2
14
-4
10
-20
20
0
x  2 x  10  0
2
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 roots
Use the given root to find the remaining roots of the function
f x   3x  5x  25x  45x  18; root  3i
4
3
2
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.
3i
-3i
3 5
25
45
-18
9i -27+15i -45-6i 18
3 5+9i -2+15i
-6i
0
-9i
-15i
6i
0
So the roots3of 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 roots.



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 Corcoran
Head of Mathematics
St Stephen’s School – Carramar
www.ststephens.wa.edu.au