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```Name ________________________________________ Date __________________ Class __________________
LESSON
4-2
Complex Numbers
Practice and Problem Solving: C
Write each imaginary number as an expression without i.
1.
6i – 9i 2
________________________
3. 2i + 5i 6
2. i 2 – i 4
_______________________
________________________
When simplified, what is the real part of the complex expression?
4. (−1 + 2i ) + (6 − 9i )
5. (3 − 3i ) − (4 + 7i )
________________________________________
________________________________________
When simplified, what is the imaginary part of the complex expression?
7. (3 + 2i)i 2
6. (4 + 5i)(2 + i )
________________________________________
________________________________________
− 3 + 7i
to answer Problems 8 and 9.
1 + 8i
(Recall that multiplication and division are inverse operations.)
Use the complex expression
8. Set up a multiplication problem that will help you simplify the complex
expression so that the denominator is free of i.
9. What is the denominator of the expression when it is
simplified? ___________________________
10. a. Does −3 i −12 = ( −3) i ( −12)? Explain.
_____________________________________________________________________________________
b. Write a general rule for the product of radicals when using
complex numbers.
_____________________________________________________________________________________
11. In a circuit, the voltage V is given by the formula V = IZ, where I is the
current and Z is the impedance. Both the current and the impedance
are represented by complex numbers. Find the voltage if the current is
2 − i3 and the impedance is 5 + 2i.
_________________________________________________________________________________________
Original content Copyright © by Houghton Mifflin Harcourt. Additions and changes to the original content are the responsibility of the instructor.
60
4. 0
c. taking square roots:
5. −3 + 3
x −3 =1
2
x2 = 4
6.
x=± 4
x = ±2
1
2
7. 1 − 2
3. −64
8. 2 + 12i
4. −9
9. −3 − 4i
5. −3
10. 7 − 11i
6. −20
11. 8 + 2i
7. −1
12. 14 + 5i
8. −
5
9
Practice and Problem Solving: C
1. i 9; 3i;
2.
2. −2
−8
−4 − 5
3.
−27 = ( −1)(27) = −1 i
i 9 i 3 =i 9 i
−6 + 9
1.
27 =
4. 5
3 = 3i 3
5. −1
6. 14
3. 3i or −3i
7. −2
4. i 2 = ( −1)2 = −1
⎛ −3 + 7i ⎞ ⎛ 1 − 8i ⎞
8. ⎜
⎟⎜
⎟
⎝ 1 + 8i ⎠ ⎝ 1 − 8 i ⎠
5. Real number; (3i )(5i ) = 15i 2 = 15(−1) =
−15
9. 65
Success for English Learners
10. a. No, because
1. It is the square root of −1. It is used to
work with negative square roots.
3i i 12i = 3 i 12i 2 = 36i 2 =
6( −1) = −6; 36 = 6
2. You could substitute the answer in the
original equation.
first multiply the radical using the
imaginary number i, and then find the
product.
3. Yes;
( ±9i )2 + 81 = 0
92 i i 2 + 81 = 0
81 i − 1 + 81 = 0
−81 + 81 = 0
11. 8 + 9i
0=0
LESSON 4.2
Practice and Problem Solving: A/B
1. 5i
2. 21i
3. −9i
Original content Copyright © by Houghton Mifflin Harcourt. Additions and changes to the original content are the responsibility of the instructor.
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