Download Recap: complex numbers

Name
November 4, 2013
Honors Advanced Mathematics
2.5/2.6 summary and extra problems page 1
Recap: complex numbers
Number system


The complex number system consists of a + bi where a and b are real numbers,
with various arithmetic operations.
The real numbers are a subset of the complex numbers: a = a + 0i corresponding to (a, 0).
So the real line is the horizontal axis of the complex plane.
Operations






To add or subtract complex numbers, just combine like terms.
To multiply complex numbers, use the distributive property and the fact i2 = –1.
To divide complex numbers, use this technique involving a conjugate:
a  bi c  di


c  di c  di
A square root of a + bi is a complex number whose square equals a + bi.
Every non-zero complex number has two square roots, which are opposites of each other.
For example, (2 – 3i)2 = –5 – 12i and (–2 + 3i)2 = –5 – 12i.
See 2.5 #60 for a problem where you’re asked to find square roots.
As of now, we don’t have a definition for how to use a complex number as an exponent.
That requires some trigonometry ideas that we haven’t established yet.
Other operations on a complex number z = a + bi:


absolute value (or modulus) | z | = a 2  b 2
conjugate
z = a – bi
Name
November 4, 2013
Honors Advanced Mathematics
2.5/2.6 summary and extra problems page 2
Summary: theorems about zeros of polynomials
Name of theorem;
location in
Precalculus text
Applies when
the coefficients
meet this
requirement
Pertains to
what kind of
zeros
What the theorem says
Remarks
Factor Theorem
p. 199
complex coeffs.
complex zeros
x = c is a zero 
(x – c) is a factor.
We proved this using the
Remainder Theorem. An
equivalent statement is that
P(x)  (x – c) has remainder
0.
Fundamental
Theorem of
Algebra
p. 210
complex coeffs.
complex zeros
P(x) of degree n
has exactly n zeros
in the complex numbers
(when counted using
their multiplicities).
Equivalently: every
polynomial can be factored
into n linear factors. The
complex number system is
called complete because it
has this complete
factorization property.
Complex
Conjugates
Theorem
p. 211
real coeffs.
complex zeros
If a + bi is a zero, then
a – bi is also a zero.
To understand why the
coefficients in this theorem
must be real, multiply out a
counterexample (ex. a + bi
is a zero but a – bi is not).
Fundamental
connection between
zeros and
x-intercepts
real coeffs.
real zeros
x = c is a zero 
the graph has an
x-intercept at x = c.
Note that non-real complex
zeros do not appear on
graphs as x-intercepts.
Intermediate Value
Theorem (applied
to finding zeros of
polynomials)
p. 190
real coeffs.
real zeros
If P(c) and P(d) have
opposite signs, then
there must be a zero
between c and d.
This is a consequence of the
fact that polynomial graphs
are continuous. The same
fact applies for any
continuous function.
Odd Degree
Theorem
p. 214
real coeffs.
real zeros
Odd degree
polynomials have
at least one real zero.
This theorem can be
understood graphically by
thinking about end behavior.
Irrational
Conjugates
Theorem
not in textbook
rational coeffs.
real zeros
If a + b is a zero
(where b is irrational)
then a – b is also a
zero.
This theorem can be thought
of as the irrational analogue
to the Complex Conjugates
Theorem.
Rational Zeros
Theorem
p. 201
integer coeffs.
rational zeros
The only possible
rational zeros are of the
of constant erm
t
form  factor
factor of leading coeff. .
Often used to prove that
certain polynomials have no
rational zeros, or to prove
that a certain zero is
irrational.
The last two theorems aren’t required knowledge for this course, but included in case you are interested.
Name
November 4, 2013
Honors Advanced Mathematics
2.5/2.6 summary and extra problems page 3
Problems about finding solutions and factors
1. Consider the equation x 3  27 .
a. How many solutions does this equation have in the real number system?
b. According to the Fundamental Theorem of Algebra, how many solutions does this equation have
in the complex number system?
c. Find all solutions of the equation in the complex number system.
2.
a. Find a quadratic polynomial with real coefficients having (–2 + 5i) as a zero. Write your answer in
standard form (that is, multiply the factors, don’t leave in factored form).
b. Find a cubic polynomial with real coefficients having 4 and (3 – 2i) as zeros. Write your answer in
standard form. Hint: Easiest if you multiply the factors from the non-real zeros first.
3. To answer these questions you will need to write polynomials that have complex numbers as
coefficients. It is not possible to write answers having real coefficients.
a. Find a linear polynomial with complex coefficients having (–2 + 5i) as a zero.
b. Find a quadratic polynomial with complex coefficients having 4 and (3 – 2i) as zeros. Write your
answer in standard form.
4. You are given this information about polynomial P(x):
 P(x) is a degree 5 polynomial with real coefficients.
 The graph of P(x) for real numbers x is given on the
grid. Its intercepts are at x = –3, x = 4, and y = 2.
 In the complex numbers, P(i+1) = 0.
a. When P(x) is divided by (x – 3), what is the remainder?
b. Factor P(x) completely in the complex number system.
c. Factor P(x) as completely as possible in the real number
system.
5. Suppose f(x) = x4 – 8x3 + 27x2 – 50x + 50.
a. Verify that f(1 – 2i) = 0.
b. Factor f(x) completely in the complex number system.
c. Factor f(x) as completely as possible in the real number system.
Name
November 4, 2013
Honors Advanced Mathematics
2.5/2.6 summary and extra problems page 4
Problems about incorrect uses of theorems
Directions for problems 6 and 7: Each of the following arguments reaches an incorrect conclusion.
Identify the erroneous step in each argument; identify which theorem was misused and explain why the
use is incorrect. The theorem summary on page 2 may be helpful. Consider the type of coefficients
required by each theorem.
1
has values f(3) = –1 and f(5) = 1. Because one of these values
x4
is negative and the other is positive, there must be a zero in between. That is, f(x) must have a zero in
the interval 3 < x < 5.”
7. Argument: “Let P be a polynomial with complex coefficients, having 2 + 3i as a zero. Since non-real
 pairs, P cannot be a polynomial of degree 1.”
zeros occur in conjugate
Follow-up question: Find polynomials of degrees 1, 2, and 3, each having 2 + 3i as a zero. Make
them polynomials with real coefficients, if possible.
6. Argument: “The function f(x) =
Problems about square roots and complex zeros
8. a. How could you show that 79 is the square root of 6421 without using a calculator?
b. How could you show that 3 – i is a square root of 8 - 6i?
2
1  i  is a square root of i.
9. Show that
2
10. Find the square roots of 3+4i. That is, find a + bi having the property (a + bi)2 = (–3 + 4i).
11. a. Find the square roots of (–13 + 8i). That is, find a + bi having the property (a + bi)2 = (–13 + 8i).
b. Check your answer by typing √(–13 + 8i) on your calculator.
12. Consider this polynomial with complex coefficients: f(x) = x2 + 3i x + (1 – 2i).
(Note that several of the theorems on page 2 apply only to polynomials with real coefficients, so do
not apply to this polynomial.)
a. Use the Quadratic Formula to find the zeros of f(x).
Hint: The result of problem 9 will be needed.
b. Write a complete factorization of f(x).
Problems about GCF and LCM
13. f(x) = x4 – 8x3 + 27x2 – 50x + 50 and g(x) = 3x5 – 26x4 + 73x3 – 44x2 – 86x + 60.
Find GCF(f(x), g(x)) and LCM(f(x), g(x)) given the hint that f(1 – 2i) = 0.
Definition: A pair of numbers or a pair of polynomials is called relatively prime if the pair has a greatest
common factor (GCF) of 1.
14. Suppose f  x  and g x  are polynomials. The zeros of f  x  are –3, 1, and 5. The zeros of g x  are –
4, –1, and 1. Are f  x  and g x  relatively prime? Explain.
15. Suppose mx   x 4  2 x 3  4 x 2  4 x  4 and n x  x 3  5x 2  8x  6 . Find all the zeros (real and
non-real) for both functions given the hint that m x  and n x  are not relatively prime.
Name
November 4, 2013
Honors Advanced Mathematics
2.5/2.6 summary and extra problems page 5
ANSWER NOTES:
Depending on the problem, what’s shown here might be a complete solution, or just the final answer, or
just a hint or a first step.
Problems about finding solutions and factors
1. c. 3,  32  3 2 3 i ,  32  3 2 3 i
2. a. x2 + 4x + 29
b. Begin from (x – 4)(x – (3 – 2i)) (x – (3 + 2i)).
The easiest start is to multiply the second and third factors.
That gives (x – 4)(x2 – 6x + 13).
Then use the distributive property to get x3 – 10x2 + 37x – 52.
3. a. x – (–2 + 5i)
b. (x – 4)(x – (3 – 2i)), then either distribute or apply the formula from 2a, to get
x2 + (–7 + 2i)x + (12 – 8i).
4. a. remainder = P(3) = 5
b.
1
36
(x + 3)2(x – 4)(x – (1 + i))(x – (1 – i))
c.
1
36
(x + 3)2(x – 4)(x2 – 2x + 2)
5. b. (x – (1 – 2i))(x – (1 + 2i))(x – (3 + i))(x – (3 – i))
c. (x2 – 6x + 10)(x2 – 2x + 5)
Problems about incorrect use of theorems
6. Intermediate Value Theorem does not apply because f(x) is not continuous on the interval.
7. Complex Conjugates Theorem does not apply when the coefficients are complex.
(x – (2 + 3i)); x2 – 4x + 13; x3 – 4x2 + 13x
Problems about square roots and complex zeros
8. a. 79 2  6421
b. 3  i   8  6i
2
2
 2

1  i   i
9. 
 2

10. 2+i and –2–i . See problem 11 notes for more hints.
Name
November 4, 2013
Honors Advanced Mathematics
2.5/2.6 summary and extra problems page 6
11. a. You need to solve the system a2 – b2 = –13, 2ab = 8. Solve the second equation for a, substitute
into the first equation. Multiply by b2 to get b4 – 13b2 – 16 = 0.
13  233
Let x = b2 and apply the quadratic formula to x2 – 13x – 16 to get x =
.
2
But x = b2 cannot be negative, so discard the negative possibility.
13  233
.
2
Perhaps the easiest way to get a is to repeat the process solving for the other variable. Get a =
Square root again to get b = 

 13  233
.
2
So a + bi =
 13  233
+
2
12. a. Quadratic formula gives
13  233
 13  233
i or –
–
2
2
 3i   13  8i
, then substitute the answers to 9a in place of the square
2
root.
Problems about GCF and LCM
13. GCF: x2 - 2x + 5; LCM: (x2 – 2x + 5)(x2 – 6x + 10)(3x3 – 8x2 –5x + 6)
14. No - because (x - 1) is a common factor for both functions.
15. m(x): x 1 i , x  i 2


13  233
i.
2
n(x): x = 3, x 1 i
