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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. Reading Strategies 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. b. Possible answer: When multiplying radicals that have negative radicands, 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. 283