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Algebra II : Basic Operations with Complex Numbers

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Example Question #41 : Basic Operations With Complex Numbers

Multiply: i(5i-6)

Possible Answers:

-5i-6i

-6i

-5-6i

-5-i

Correct answer:

-5-6i

Explanation:

Use the distributive property to multiply:

5i^2-6i .

Recall that

i^2=-1

so that you get

5(-1)-6i ,

which gives you a final answer of

-5-6i .

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Example Question #42 : Basic Operations With Complex Numbers

Evaluate: 9i^4-3i^2

Possible Answers:

Correct answer:

Explanation:

In order to solve this, we will need to convert the imaginary terms.

Recall that $i=\sqrt{-1}$ .

i^2 =-1

$i^4 = i^2\cdot i^2 = 1$

Replace the values with the imaginary terms.

9i^4-3i^2 = 9(1)-3(-1) = 9+3 = 12

The answer is:

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Example Question #43 : Basic Operations With Complex Numbers

Simplify: $\sqrt{-9}+ \sqrt9 +\sqrt{-1}$

Possible Answers:

$4i\times3$

$2i\sqrt2+3$

$\textup{Cannot be simplified.}$

4i+3

10i+3

Correct answer:

4i+3

Explanation:

The negative terms inside the square roots will be imaginary.

The term, , is the imaginary term for $\sqrt{-1}$ .

Rewrite the terms using this as the common factor so that we can replace $\sqrt{-1}$ with .

$\sqrt{-9}+ \sqrt9 +\sqrt{-1} = \sqrt{-1}\cdot \sqrt{9}+ \sqrt{9}+\sqrt{-1}$

Replace the terms and simplify the square roots.

$\sqrt{-1}\cdot \sqrt{9}+ \sqrt{9}+\sqrt{-1} = i(3)+3+i = 4i\pm3$

The answer is: 4i+3

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Example Question #44 : Basic Operations With Complex Numbers

Solve: $i^{2}-i^{18}$

Possible Answers:

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

Correct answer:

Explanation:

Evaluate each imaginary term. The value of $i = \sqrt{-1}$ .

$i^2 = i\cdot i = \sqrt{-1} \cdot \sqrt{-1} = -1$

Evaluate $i^{18}$ . Rewrite the exponent as a product of exponents.

$i^{18} =( i^2)^9 = (-1)^9 = -1$

Combine and subtract the two terms.

$i^{2}-i^{18} = -1-(-1) = -1+1 =0$

The answer is:

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Example Question #45 : Basic Operations With Complex Numbers

(5i-2)(i+3)

Possible Answers:

13i-11

13i-1

13-11i

13i

Correct answer:

13i-11

Explanation:

This problem requires you to use FOIL to multiply the binomials.

Multiply the first terms

$(5i\cdot i=5i^2)$ ,

then the outside terms

$(5i\cdot 3=15i)$ ,

next the inside terms

$(-2\cdot i=-2i)$ ,

and finally the last terms

$(-2\cdot 3=-6)$ .

Put those together to get

5i^2+15i-2i-6=5i^2+13i-6 .

Recall that

i^2=-1 .

Therefore, your answer is

13i-11 .

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Example Question #46 : Basic Operations With Complex Numbers

Simplify: $-2i^{2}-4i^{50}+8i^{100}$

Possible Answers:

10-4i

10+4i

6+4i

Correct answer:

Explanation:

Recall that $i=\sqrt{-1}$ . We can find the powers of this imaginary term by writing out some of the powers.

$i^2 = i\cdot i = \sqrt{-1}\cdot \sqrt{-1} = -1$

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

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

$i^5 = i^4 \cdot i =i$

Notice that this pattern will repeat.

Convert each term of $-2i^{2}-4i^{50}+8i^{100}$ to its proper value.

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

$4i^{50} = 4(i^2)^{25} = 4(-1)^{25} = 4(-1) = -4$

$8i^{100} = 8(i^4)^{25} = 8(1)^{25} = 8$

Rewrite the expression.

$-2i^{2}-4i^{50}+8i^{100} = 2-(-4)+8 = 2+4+8 = 14$

The answer is:

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Example Question #41 : Basic Operations With Complex Numbers

$What\ is\ the\ complex\ conjugate\ for\ 9-2i?$

Possible Answers:

$81-4i^{2}$

$81+4i^{2}$

9-2i

9+2i

Correct answer:

9+2i

Explanation:

$The\ complex\ conjugate\ for\ a binomial\ with\ an\ imaginary\ number\ is$

$simply\ the\ original\ with\ the\ sign\ for\ the\ imaginary\ number\ changed.$

$If\ 9-2i\ were\ multiplied\ by\ 9+2i\ the\ i\ values\ would\ cancel\ producing\ a\ real\ number.$

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Example Question #42 : Basic Operations With Complex Numbers

Simplify: (17+13i)+(8-10i)

Possible Answers:

9+3i

266-66i

25+23i

25+3i

Correct answer:

25+3i

Explanation:

Simplify: (17+13i)+(8-10i)

$Drop\ the\ ()\ and\ combine\ like\ terms\ 17+13i+8-10i$

17+8+13i-10i=25+3i

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Example Question #43 : Basic Operations With Complex Numbers

$Simplify:\ (3-5i)-(2+7i)$

Possible Answers:

1+2i

-41-11i

1-12i

41+11i

Correct answer:

1-12i

Explanation:

When combining imaginary numbers we must first distribute the negative sign.

(3-5i)-(2+7i)=3-5i-2-7i

Now combine like terms.

3-2-5i-7i=1-12i

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Example Question #50 : Basic Operations With Complex Numbers

Simplify the term: $i^{20}-i^{25}$

Possible Answers:

-1-i

i+1

1-i

-2i

Correct answer:

1-i

Explanation:

Write the value of the imaginary value.

$i=\sqrt{-1}$

Determine the values of the first several powers of .

$i^2 = i\cdot= \sqrt{-1}\cdot \sqrt{-1} = -1$

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

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

$i^5 = i\cdot i^4= i$

Notice that this pattern will repeat.

Rewrite the expression in terms of the product of powers.

$i^{20}-i^{25} =( i^4)^5- i(i^{24}) = (1)^5-i((i^4)^6)$

Simplify i((i^4)^6) .

i((i^4)^6) = i((1)^6)=i

Replace the term.

(1)^5-i((i^4)^6) = 1-i

The answer is: 1-i

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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 #41 : Basic Operations With Complex Numbers

Example Question #42 : Basic Operations With Complex Numbers

Example Question #43 : Basic Operations With Complex Numbers

Example Question #44 : Basic Operations With Complex Numbers

Example Question #45 : Basic Operations With Complex Numbers

Example Question #46 : Basic Operations With Complex Numbers

Example Question #41 : Basic Operations With Complex Numbers

Example Question #42 : Basic Operations With Complex Numbers

Example Question #43 : Basic Operations With Complex Numbers

Example Question #50 : Basic Operations With Complex Numbers

All Algebra II Resources

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