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Professor P. Bishop
MAC1105 (MDN)
1.3 Complex Numbers; Quadratic Equations with a Negative Discriminant
The complex number system enables us to take even roots of negative numbers by means of the imaginary unit
i, which is equal to the square root of –1; that is i2 = -1 and i = 1 . By factoring –1 out of a negative
expression, it becomes positive and an even root can be taken: -b = i b . Standard form for complex
expression is a + bi, where a is the real part and bi is the imaginary part. All properties of exponents hold when
the base is i, thus i1 = i, i2 = -1, i3 = i2(i) = -1i = -i, i4 = i2(i2) = -1(-1) = 1. In general, for in, divide n by 4: if the
remainder is 0, in= 1; if the remainder is 1, in = i, if the remainder is 2, in= -1; if the remainder is 3, in= -i. The
product of a complex number (a + bi) and its conjugate (a – bi) is a nonnegative real number (a2 + b2).
Write in a + bi form:
1. 25
2. a.  50
b. 50
Evaluate (simplify to 1, -1, i, -i):
4. i21
5. i39
6. i52
Write in a + bi form:
8. (-1 + 6i) + (5 – 4i)
9
10. 2  10
15.
4i10
+
6i5
7. i42
(-6 + 4i) - (2 – i) + (7 – 3i)
11. -5i7(2 + 3i)
13. (4 + 3 i)3
12. (3 – 2 i)(2 – 5i)
14.
+
3i4
+
5i3
3. 3 16  2
16.
(2 + 6i)(2 – 6i)
8
2
17.
3  5i
2i
18.
5  6i
6  3i
Solve in the complex number system:
19. x2 + 49 = 0
20. 5x2 + 4x + 8 = 0
21. x3 – 64 = 0
22. x4 + 3x2 – 4 = 0
23.
x3 + 4x2 – 4x – 16 = 0
TRY THESE:
24. Write in a + bi form: 2i7(5 + 3i3)(6 – 2i9)
25. Solve in the complex number system: 2x4 – 112 = 25z2
Professor P. Bishop
MAC1105 (MDN)
5.4 COMPLEX ZEROS; FUNDAMENTAL THEOREM OF ALGEBRA
Fundamental Theorem of Algebra: Every complex polynomial function of degree n ≥ 1 or higher has at least
one complex zero. Theorem: Every complex polynomial of degree n ≥ or higher can be factored into n linear
factors (not necessarily distinct). Conjugate Pairs Theorem: Given a polynomial with real coefficients, if r =
a + bi is a zero, then r  a  bi is also a zero of the polynomial.
Information is given about a polynomial f(x) whose coefficients are real numbers. Find the remaining zeros:
1. Degree 4; zeros: 3, 5, 4 + i
2. Degree 7; zeros: 2, 3i, 5 – i, 3 + i
Form a polynomial f(x) with real coefficients having the given degree and zeros:
3. Degree 4; zeros: 4 – 3i, -3 multiplicity 2
4. Degree 5; zeros: 3 multiplicity 3, 2 – i
Use the given zero to find the remaining zeros:
5. f(x) = x4 – 7x3 + 14x2 – 38x + 60; zero: 1 + 3i
Find the complex zeros of the polynomial function. Write f in factored form:
6. f(x) = 2x4 + x3 – 35x2 – 113x + 65
Max. Zeros:
Potential Zeros:
Real Zeros:
Positive Zeros:
Negative Zeros:
Try These (5.4)
Pg. 386:
#24.
#32.