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College Algebra : Complex Numbers

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Example Questions

Example Question #1 : Complex Numbers

Consider the following definitions of imaginary numbers:

x = 4 - 2i

y = 6 +7i

z = 5i

Then, x+y-2z = ?

Possible Answers:

10-i

2(5+i)

5(2-i)

-5i

Correct answer:

5(2-i)

Explanation:

x + y-2z = (4-2i) + (6+7i) -2(5i) = 10 - 5i = 5(2-i)

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Example Question #1 : Complex Numbers

What is the value of (3 + i)(4 + i) ?

Possible Answers:

12 + 8i

11 + 7i

12 + 6i

Correct answer:

11 + 7i

Explanation:

When dealing with imaginary numbers, we multiply by foiling as we do with binomials. When we do this we get the expression below:

(3+i)(4+i)= 12 + 3i + 4i + i^2

Since we know that i^2 = -1 we get 12 + 7i - 1 which gives us 11 + 7i .

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Example Question #1 : Complex Numbers

What is the value of i^4 ?

Possible Answers:

i + 1

Correct answer:

Explanation:

Recall that the definition of imaginary numbers gives that $i = \sqrt{-1}$ and thus that i^2 = -1 . Therefore, we can use Exponent Rules to write $i^4 = i^2 \cdot i^2 = -1\cdot-1 = 1$

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Example Question #2 : Complex Numbers

Add:

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

Possible Answers:

5-3i

1-5i

5+3i

-1+5i

1+5i

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 #2 : Complex Numbers

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

The answer must be in standard form.

Possible Answers:

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

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

$\frac{1}{5i}$

$\frac{2}{3}i$

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:

2i-3

3+2i

3-2i

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 #1 : Complex Numbers

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

Possible Answers:

-3+i

1-3i

-1-3i

1+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 #2 : Complex Numbers

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

Possible Answers:

8-6i

7(1-i)

The answer is not present.

7-6i

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 #12 : Basic Operations With Complex Numbers

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

Possible Answers:

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

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

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

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

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

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 #1 : Complex Numbers

Multiply: (3+i)(5-4i)

Possible Answers:

19-7i

11-7i

19+17i

15-3i

15-11i

Correct answer:

19-7i

Explanation:

Use FOIL to multiply the two binomials.

Recall that FOIL stands for Firsts, Outers, Inners, and Lasts.

(3+i)(5-4i)

15+5i-12i-4i^2

Remember that i^2=-1

15-7i+4

19-7i

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College Algebra : Complex Numbers

Study concepts, example questions & explanations for College Algebra

All College Algebra Resources

Example Questions

Example Question #1 : Complex Numbers

Example Question #1 : Complex Numbers

Example Question #1 : Complex Numbers

Example Question #2 : Complex Numbers

Example Question #2 : Complex Numbers

Example Question #1 : Complex Numbers

Example Question #1 : Complex Numbers

Example Question #2 : Complex Numbers

Example Question #12 : Basic Operations With Complex Numbers

Example Question #1 : Complex Numbers

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