Download Review: complex numbers

Name
October 25/26, 2012
Honors Advanced Mathematics
2.5 extra problems page 1
Review: 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.
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.

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 (we’ll work with this later in the year.)
conjugate
z = a – bi
Name
October 25/26, 2012
Honors Advanced Mathematics
2.5 extra problems page 2
Part I: complex numbers
1. Show that the sum of a + bi and its conjugate is a real number.
2. Show that the product a + bi and its conjugate is a nonnegative real number.
3a. How could you show that 79 is the square root of 6421 without using a calculator?
3b. How could you show that 3 – i is a square root of 8 - 6i?
4. Show that 4-3i is a square root of 7-24i.
5. Show that
2
1  i  is a square root of i.
2
6. Find the square roots of 3+4i.
Name
October 25/26, 2012
Honors Advanced Mathematics
2.5 extra problems page 3
Write the following in the form a + bi.
7.
1
2  5i
8.
9.
3i 2
7i 2
10.
11. i  i 2  i 3  i 4  i 5
1
4  3i
2i 5
3i 5
12. i 46  i 47
Part II More Practice with 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
October 25/26, 2012
Honors Advanced Mathematics
2.5 extra problems page 4
Part III:
Problems about solutions and factors of polynomial
functions
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
October 25/26, 2012
Honors Advanced Mathematics
2.5 extra problems page 5
Problems about incorrect uses of theorems
Directions: The following argument reaches an incorrect conclusion. Identify the erroneous step
in the argument, identifying which theorem was misused and explaining why the use is incorrect.
Then, answer the follow-up question.
6. Argument: “Let P be a polynomial with complex coefficients, having 2 + 3i as a zero. Since
non-real zeros occur in conjugate pairs, P cannot be a polynomial of degree 1.”
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.
Problems about square roots and complex zeros
7. 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.
8. Consider this polynomial with complex coefficients: f(x) = x2 + 3i x + (1 – 2i).
a. Use the Quadratic Formula to find the zeros of f(x).
Hint: The result of problem 7 will be needed.
b. Write a complete factorization of f(x).
ANSWERS:
Parts I and II:
2 . a 2  b2  0
1. 2a
3. a. 79 2  6421
4. 4  3i   7  24i
b. 3  i   8  6i
2
2
2
 2

1  i   i
5. 
 2

9.
19 10 2

i
51
51
6. 2+i and –2–i
10.
1 5 5

i
14 14
11. i
7.
2
5
 i
29 29
8.
12. –1–i
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
n(x): x = 3, x 1 i



4
3
 i
25 25
Name
October 25/26, 2012
Honors Advanced Mathematics
2.5 extra problems page 6
Part III
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. We’ll go over questions on Monday. I can also answer give
some help by email when I am away on Thursday and Friday.
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)
6. Complex Conjugates Theorem does not apply when the coefficients are complex.
(x – (2 + 3i)); x2 – 4x + 13; x3 – 4x2 + 13x
7. 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.
Square root again to get b = 
Get a = 
So a + bi =
 13  233
.
2
 13  233
+
2
8. a. Quadratic formula gives
the square root.
13  233
 13  233
i or –
–
2
2
13  233
i.
2
 3i   13  8i
, then substitute the answers to 7a in place of
2