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Algebra II : Basic Operations with Complex Numbers

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Example Questions

Example Question #61 : Basic Operations With Complex Numbers

Add: 2i^2+9i^4-3i^8

Possible Answers:

Correct answer:

Explanation:

Write the imaginary values of the powers given. Recall that $i=\sqrt{-1}$ .

$i^2= i\cdot i = -1$

$i^4 = i^2\cdot i^2 = (-1)\cdot (-1) = 1$

$i^8 = i^4\cdot i^4 = 1\cdot 1 =1$

Substitute the values into the expression.

2i^2+9i^4-3i^8 = 2(-1)+9(1)-3(1) = -2+9-3

The answer is:

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Example Question #2071 : Mathematical Relationships And Basic Graphs

Evaluate: $-(i^{-36})$

Possible Answers:

2-i

-2i

Correct answer:

Explanation:

Rewrite the expression without the negative exponent.

$x^{-a} = \frac{1}{x^a}$

$-(i^{-36}) = -(\frac{1}{i^{36}})=-\frac{1}{i^{36}}$

Write out the first few terms of the imaginary number.

$i=\sqrt{-1}$

$i^2= i\cdot i = -1$

$i^3 = i^2\cdot i = -i$

$i^4 = i^2\cdot i^2 = (-1)\cdot (-1) = 1$

This will repeat for higher powers.

The power of 36 goes into 4 nine times. We can rewrite the exponent as a product of exponents.

$-\frac{1}{i^{36}} = -\frac{1}{(i^4)^9}$

Replace i^4 with the known value.

$-\frac{1}{(i^4)^9} = -\frac{1}{(1)^9} = -1$

The answer is:

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Example Question #63 : Basic Operations With Complex Numbers

Solve: $[(i^2)^3]^{-4}$

Possible Answers:

$\textup{The answer is not given.}$

Correct answer:

Explanation:

Write the values of the first two imaginary terms.

$i=\sqrt{-1}$

$i^2= i\cdot i = -1$

Replace the term inside the expression.

$[(-1)^3]^{-4} = [-1]^{-4}$

Rewrite the negative exponent into a fraction.

$\frac{1}{(-1)^4} = \frac{1}{1} = 1$

The answer is:

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Example Question #4741 : Algebra Ii

Solve: $-2i^2-4i^{-4}+6i^6$

Possible Answers:

Correct answer:

Explanation:

Evaluate each term of the expression. Write out the values of the imaginary terms.

$i=\sqrt{-1}$

$i^2= i\cdot i = -1$

$i^4 = i^2\cdot i^2 = (-1)\cdot (-1) = 1$

$i^6 = i^4 \cdot i^2 = (1)\cdot (-1) = -1$

Replace the values of each.

-2i^2 = -2(-1) = 2

$-4i^{-4} = \frac{-4}{i^4} = -\frac{4}{1} = -4$

6i^6 = 6(-1) =-6

Sum all the values.

$-2i^2-4i^{-4}+6i^6 = 2-4-6 = -8$

The answer is:

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Example Question #65 : Basic Operations With Complex Numbers

Solve: -i^2-2i^4-15i^6

Possible Answers:

Correct answer:

Explanation:

Write out the terms of the imaginary powers.

$i=\sqrt{-1}$

$i^2= i\cdot i = -1$

$i^4 = i^2\cdot i^2 = (-1)\cdot (-1) = 1$

$i^6=i^4\cdot i^2 = (1)\cdot (-1) = -1$

Replace all terms with the correct value.

-i^2-2i^4-15i^6=-(-1)-2(1)-15(-1)

1-2+15= 14

The answer is:

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Example Question #66 : Basic Operations With Complex Numbers

Add the following terms: $i^6-i^4-3i^2-i^{10}$

Possible Answers:

Correct answer:

Explanation:

Simplify each term. Recall that:

$i=\sqrt{-1}$

$i^2= i\cdot i = -1$

$i^4 = i^2\cdot i^2 = (-1)\cdot (-1) = 1$

This means that:

$i^6= i^4\cdot i^2 = (1)\cdot (-1) = -1$

$i^{10} =( i^2)^5 = (-1)^{5} = -1$

Subtract the terms.

$i^6-i^4-3i^2-i^{10}=-1-1-3(-1)-(-1)$

Simplify the terms.

-2+3+1 = 2

The answer is:

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Example Question #67 : Basic Operations With Complex Numbers

Solve: $\frac{2i^3-9i^5}{i}$

Possible Answers:

9i+2

-9i+2

Correct answer:

Explanation:

Simplify by dividing the imaginary term.

$\frac{2i^3-9i^5}{i} = 2i^2-9i^4$

Write the first few terms of the imaginary powers.

$i=\sqrt{-1}$

$i^2= i\cdot i = -1$

$i^4 = i^2\cdot i^2 = (-1)\cdot (-1) = 1$

Substitute the values into the expression.

2i^2-9i^4 = 2(-1)-9(1) = -2-9 = -11

The answer is:

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Example Question #61 : Basic Operations With Complex Numbers

(6-3i)-(3+6i)

Possible Answers:

9-3i

-9i-3

None of these.

3+9i

3-9i

Correct answer:

3-9i

Explanation:

(6-3i)-(3+6i)

Distribute the negative sign:

6-3i-3-6i

Combine like terms:

$\mathbf{3-9i}$

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Example Question #4743 : Algebra Ii

Evaluate: (i^2+2)(i+3)(i^8-4)

Possible Answers:

-9i+27

-5i-15

-5i+15

-3i-9

-3i+9

Correct answer:

-3i-9

Explanation:

Recall that $i=\sqrt{-1}$ . We can then write the terms for the powers of the imaginary term.

$i^2= i\cdot i = -1$

$i^4 = i^2\cdot i^2 = (-1)\cdot (-1) = 1$

This means that, using the distribution property of exponents:

i^8 =( i^4)^2 = (1)^2 = 1

Replace the terms.

(i^2+2)(i+3)(i^8-4)=(-1 +2)(i+3)(1-4)

= (1)(i+3)(-3) = -3(i+3)

Distribute the integer through the binomial.

The answer is: -3i-9

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Example Question #63 : Basic Operations With Complex Numbers

Evaluate: i^4-i^9-i

Possible Answers:

-2i-1

-2i+1

-2i-2

Correct answer:

-2i+1

Explanation:

Write the powers of the imaginary numbers.

$i=\sqrt{-1}$

$i^2= i\cdot i = -1$

$i^3 = i^2\cdot i = -i$

$i^4 = i^2\cdot i^2 = (-1)\cdot (-1) = 1$

Notice that this will repeat. We can rewrite higher powers if the imaginary term by product of powers.

i^4-i^9-i= 1-i(i^4)^2-i=1-i-i= -2i+1

The answer is: -2i+1

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Algebra II : Basic Operations with Complex Numbers

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #61 : Basic Operations With Complex Numbers

Example Question #2071 : Mathematical Relationships And Basic Graphs

Example Question #63 : Basic Operations With Complex Numbers

Example Question #4741 : Algebra Ii

Example Question #65 : Basic Operations With Complex Numbers

Example Question #66 : Basic Operations With Complex Numbers

Example Question #67 : Basic Operations With Complex Numbers

Example Question #61 : Basic Operations With Complex Numbers

Example Question #4743 : Algebra Ii

Example Question #63 : Basic Operations With Complex Numbers

All Algebra II Resources

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