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Algebra II : Imaginary Numbers

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Example Questions

Example Question #4 : Basic Operations With Complex Numbers

Add:

(2-3i)+(-1-2i)

Possible Answers:

1+5i

-1+5i

1-5i

5+3i

5-3i

Correct answer:

1-5i

Explanation:

When adding complex numbers, add the real parts and the imaginary parts separately to get another complex number in standard form.

Adding the real parts gives 2-1=1 , and adding the imaginary parts gives -5i .

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Example Question #5 : Basic Operations With Complex Numbers

Divide: $\frac{2-i}{3+2i}$

The answer must be in standard form.

Possible Answers:

$\frac{2}{3}i$

$\frac{2}{3}-\frac{1}{2}i$

$\frac{1}{5i}$

$\frac{4}{13}-\frac{7i}{13}$

Correct answer:

$\frac{4}{13}-\frac{7i}{13}$

Explanation:

Multiply both the numerator and the denominator by the conjugate of the denominator which is 3-2i which results in

$\frac{\left ( 2-i \right )\left ( 3-2i \right )}{\left ( 3+2i \right )\left ( 3-2i \right )}$

The numerator after simplification give us $6-4i-3i+2i^{2}=6-7i+2i^{2}=4-7i$

The denominator is equal to $3^{2}-4i^{2}=9+4=13$

Hence, the final answer in standard form =

$\frac{4}{13}-\frac{7i}{13}$

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Example Question #1 : Complex Numbers

Divide: $\frac{2+3i}{i}$

Answer must be in standard form.

Possible Answers:

3+2i

3-2i

2i-3

Correct answer:

3-2i

Explanation:

Multiply both the numerator and the denominator by the conjugate of the denominator which is resulting in

$\frac{\left ( 2+3i \right )\left ( -i \right )}{\left ( i \right )\left ( -i \right )}$

This is equal to $\frac{-2i-{3i}^2}{-i^2}$

Since i^2 = -1 you can make that substitution of in place of i^2 in both numerator and denominator, leaving:

$\frac{-2i-3(-1)}{-(-1)}$

When you then cancel the negatives in both numerator and denominator (remember that -(-1)=1 , simplifying each term), you're left with a denominator of and a numerator of -2i + 3 , which equals 3-2i .

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Example Question #2 : Complex Numbers

Evaluate: i(1+i)(2-i)

Possible Answers:

1+3 i

-1-3i

-1+3i

-3+i

1-3i

Correct answer:

-1+3i

Explanation:

Use the FOIL method to simplify. FOIL means to mulitply the first terms together, then multiply the outer terms together, then multiply the inner terms togethers, and lastly, mulitply the last terms together.

i(1+i)(2-i)

=i[(1)(2)+(1)(-i)+(i)(2)+(i)(-i)]

=i[2-i+2i-i^2]

=i[2+i-i^2]

=2i+i^2-i^3

The imaginary is equal to:

$\sqrt{-1}$

Write the terms for $i, i^2, \textup{and } i^3$ .

$i=\sqrt{-1}$

i^2=-1

$i^3=i \times i^2=-\sqrt{-1}=-i$

Replace $i^2 \textup{ and } i^3$ with the appropiate values and simplify.

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

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Example Question #73 : Imaginary Numbers

What is the value of $i^{42}$ , if $\small i$ = $\small \sqrt{-1}$ ?

Possible Answers:

$\textup{No solution}$

$\small i$

$\small -i$

Correct answer:

Explanation:

We know that $\small i=\sqrt{-1}}$ . Therefore, $\small i^{2}=-1, i^{3}=-\sqrt{-1}, i^{4}=1$ . Thus, every exponent of $\small i$ that is a multiple of 4 will yield the value of $\small 1$ . This makes $\small i^{40} =1$ . Since $\small i^{42}=i^{40}*i*i$ , we know that $\small i^{42}=1*i*i=i^{2}=-1$ .

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Example Question #3 : Complex Numbers

(3+2i)+4-i-i(7+i)

Possible Answers:

7(1-i)

The answer is not present.

7-6i

8-6i

Correct answer:

8-6i

Explanation:

(3+2i)+4-i-i(7+i)

Combine like terms:

7+i-i(7+i)

Distribute:

7+i-7i-i^2

Combine like terms:

7+i-7i-(-1)

$\mathbf{8-6i}$

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Example Question #13 : Basic Operations With Complex Numbers

$\sqrt{-4}*\sqrt{-25}*\sqrt{-64}$

Possible Answers:

80i

80i^3

-8i

-80i

Correct answer:

-80i

Explanation:

$\sqrt{-4}*\sqrt{-25}*\sqrt{-64}$

2i*5i*8i=80i^3

i^3=-1

$\mathbf{-80i}$

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Example Question #4 : Complex Numbers

Rationalize the complex fraction: $\frac{2+i}{3-i}$

Possible Answers:

$\frac{1-i}{2}$

$\frac{5+i}{10}$

$\frac{7+5i}{10}$

$\frac{5+5i}{8}$

$\frac{1+i}{2}$

Correct answer:

$\frac{1+i}{2}$

Explanation:

To rationalize a complex fraction, multiply numerator and denominator by the conjugate of the denominator.

$\frac{(2+i)(3+i)}{(3-i)(3+i)}$

$\frac{6+3i+2i+i^2}{9-3i+3i-i^2}$

$\frac{5+5i}{10}$

$\frac{1+i}{2}$

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Example Question #4 : Complex Numbers

Rationalize the complex fraction: $\frac{2+i}{3-i}$

Possible Answers:

$\frac{1-i}{2}$

$\frac{5+i}{10}$

$\frac{7+5i}{10}$

$\frac{5+5i}{8}$

$\frac{1+i}{2}$

Correct answer:

$\frac{1+i}{2}$

Explanation:

To rationalize a complex fraction, multiply numerator and denominator by the conjugate of the denominator.

$\frac{(2+i)(3+i)}{(3-i)(3+i)}$

$\frac{6+3i+2i+i^2}{9-3i+3i-i^2}$

$\frac{5+5i}{10}$

$\frac{1+i}{2}$

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Example Question #2031 : Mathematical Relationships And Basic Graphs

Multiply: (3i + 1 )(3 - 4i)

Possible Answers:

3-7i

2 - 12i

5i-9

5i+15

Correct answer:

5i+15

Explanation:

Distribute:

3i(3-4i) = 9i + 12

1(3-4i) = 3 - 4i

combine like terms:

5i + 15

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Algebra II : Imaginary Numbers

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #4 : Basic Operations With Complex Numbers

Example Question #5 : Basic Operations With Complex Numbers

Example Question #1 : Complex Numbers

Example Question #2 : Complex Numbers

Example Question #73 : Imaginary Numbers

Example Question #3 : Complex Numbers

Example Question #13 : Basic Operations With Complex Numbers

Example Question #4 : Complex Numbers

Example Question #4 : Complex Numbers

Example Question #2031 : Mathematical Relationships And Basic Graphs

All Algebra II Resources

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