AnswersCR - Algebra 2 26-27 - S1Operations with Complex Numbers

Operations with Complex Numbers Answers

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Aiko is finding the sum (4 + 5i) + (–3 + 7i). She rewrites the sum as (–3 + 7)i + (4 + 5)i. Which statement explains the error Aiko made by using a mathematical property incorrectly?

A
Aiko incorrectly used the commutative property by changing the order of the two complex numbers.
B
Aiko incorrectly used the associative property by changing the order of the two complex numbers.
C
Aiko incorrectly used the identity property by combining the real number and the coefficient of the imaginary part.
D
Aiko incorrectly used the distributive property by combining the real number and the coefficient of the imaginary part.
2
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Which equation illustrates the identity property of multiplication?

A
(a + bi) × c = (ac + bci)
B
(a + bi) × 0 = 0
C
(a + bi) × (c + di) = (c + di) × (a + bi)
D
(a + bi) × 1 = (a + bi)
3

Which property of addition is shown in the equation below?a + bi + 0 + 0i = a + bi

A
commutative property
B
inverse property
C
identity property
D
associative property
4

What is the additive inverse of the complex number –8 + 3i?

A
–8 – 3i
B
–8 + 3i
C
8 – 3i
D
8 + 3i
5

What is the additive inverse of the complex number 9 – 4i?

A
–9 – 4i
B
–9 + 4i
C
9 – 4i
D
9 + 4i
6

What is the value of (–7 + 3i) – (2 – 6i)?

A
–9 + 9i
B
–9 – 3i
C
–5 – 3i
D
–5 + 9i
7

What is the value of the product (3 – 2i)(3 + 2i)?

A
5
B
9 + 4i
C
9 – 4i
D
13
8

Evaluate when and .

Question illustration
A
29 i
Option A
B
29 i minus 20
Option B
C
Negative 21 i
Option C
D
29
Option D
9

Donte simplified the expression below.What mistake did Donte make?

Question illustration
A
He did not apply the distributive property correctly for 4(1 + 3i).
B
He did not distribute the subtraction sign correctly for 8 – 5i.
C
He added the real number and coefficient of i in 4(1 + 3i).
D
He added the two complex numbers instead of subtracted.
10

Which equation shows an example of the associative property of addition?

A
(–4 + i) + 4i = –4 + (i + 4i)
B
(–4 + i) + 4i = 4i + (–4i + i)
C
4i × (–4i + i) = (4i – 4i) + (4i × i)
D
(–4i + i) + 0 = (–4i + i)

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