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High School Math : High School Math

Study concepts, example questions & explanations for High School Math

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All High School Math Resources

8 Diagnostic Tests 613 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #6 : Imaginary Numbers

Evaluate: $(4 + 3i)^{2}$

Possible Answers:

7 + 24i

19 + 24i

13 - 24i

Correct answer:

7 + 24i

Explanation:

$(4 + 3i)^{2}$

$= 4^{2} + 2 \cdot 4 \cdot 3i +\left ( 3i \right )^{2}$

= 16 + 24i + 9 (-1)

= 16 + 24i -9

= 7 + 24i

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Example Question #1221 : High School Math

What is the absolute value of 4-3i

Possible Answers:

$\frac{-4}{3}$

$\frac{3}{4}$

Correct answer:

Explanation:

The absolute value is a measure of the distance of a point from the origin. Using the pythagorean distance formula to calculate this distance.

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Example Question #5 : Imaginary Numbers

Which of the following is equivalent to:

$-4i^{4} ?$

Possible Answers:

-4i

Correct answer:

Explanation:

Recall that $i^{2} = -1$ .

Then, we have that $-4(i^{4}) = -4(i^{2})^{2} = -4(-1)^{2} = -4(1) = -4$ .

Note that we used the power rule of exponents and the order of operations to simplify the exponent before multiplying by the coefficient.

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Example Question #2021 : Mathematical Relationships And Basic Graphs

Simplify the expression.

4i^2-6i-7i^2+3i+4

Possible Answers:

7-3i

-3i^2-3i+4

None of the other answer choices are correct.

-7-3i

Correct answer:

7-3i

Explanation:

Combine like terms. Treat $\small i$ as if it were any other variable.

4i^2-6i-7i^2+3i+4

-3i^2-3i+4

Substitute to eliminate $\small i^2$ .

i^2=-1

-3(-1)-3i+4

Simplify.

3-3i+4=7-3i

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Example Question #1 : Imaginary Numbers

Which of the following is equivalent to $5i^{2}$ ?

Possible Answers:

None of the other answer choices are correct.

Correct answer:

Explanation:

Recall the basic property of imaginary numbers, $i^{2} = -1$ .

Keeping this in mind, $5i^{2} = 5(-1) = -5$ .

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Example Question #1222 : High School Math

Simplify $\frac{4}{3+2i}$

Possible Answers:

$\frac{12-3i}{5i}$

$\frac{12-8i}{5}$

$\frac{4}{3}+2i$

$\frac{12-8i}{13}$

$\frac{12+8i}{5}$

Correct answer:

$\frac{12-8i}{13}$

Explanation:

Multiplying top and bottom by the complex conjugate (3-2i) eliminates i from the denominator

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Example Question #2 : Using Expressions With Complex Numbers

Multiply, then simplify.

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

Possible Answers:

6+6i

12+6i-6i^2

18+6i

Correct answer:

18+6i

Explanation:

Use FOIL to multiply. We can treat $\small i$ as a variable for now, just as if it were an $\small x$ .

(3+3i)(4-2i)=12-6i+12i-6i^2

Combine like terms.

12+6i-6i^2

Now we can substitute for $\small i^2$ .

i^2=-1

12+6i-6(-1)

Simplify.

12+6i+6=18+6i

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Example Question #1 : Using Expressions With Complex Numbers

Which of the following is equivalent to $\frac{3+i}{3-4i}$ , where $i=\sqrt{-1}$ ?

Possible Answers:

5-15i

$\frac{3-i}{3-4i}$

$\frac{1}{5}+\frac{3}{5}i$

$\frac{3+i}{3+4i}$

$\frac{1}{5}-\frac{3}{5}i$

Correct answer:

$\frac{1}{5}+\frac{3}{5}i$

Explanation:

We will need to simplify the rational expression by removing imaginary terms from the denominator. Often in such problems, we want to multiply the numerator and denominator by the conjugate of the denominator, which will usually eliminate the imaginary term from the denominator.

In this problem, the denominator is 3-4i . Remember that, in general, the conjugate of the complex number a+bi is equal to a-bi , where a and b are both nonzero constants. Thus, the conjugate of 3-4i is equal to 3+4i .

We need to multiply both the numerator and denominator of the fraction $\frac{3+i}{3-4i}$ by 3+4i .

$\frac{3+i}{3-4i}=\frac{3+i}{3-4i}\cdot\frac{3+4i}{3+4i}=\frac{(3+i)(3+4i)}{(3-4i)(3+4i)}$

Next, we use the FOIL method to simplify both the numerator and denominator. The FOIL method of binomial multiplication requires us to mutiply together the first, outside, inner, and last terms of each binomial and then sum them.

$\frac{(3+i)(3+4i)}{(3-4i)(3+4i)}=\frac{3(3)+3(4i)+i(3)+i(4i)}{3(3)+3(4i)-4i(3)+(-4i)(4i)}$

Now, we can start simplifying.

$\frac{3(3)+3(4i)+i(3)+i(4i)}{3(3)+3(4i)-4i(3)+(-4i)(4i)}=\frac{9+12i+3i+4i^2}{9+12i-12i-16i^2}$

Use the fact that $i^2=(\sqrt{-1})^2=-1$ .

$\frac{9+12i+3i+4i^2}{9+12i-12i-16i^2}=\frac{9+12i+3i+4(-1)}{9+12i-12i-16(-1)}$

$=\frac{9+15i-4}{9+16}=\frac{5+15i}{25}=\frac{5}{25}+\frac{15i}{25}$

$=\frac{1}{5}+\frac{3}{5}i$

The answer is $\frac{1}{5}+\frac{3}{5}i$ .

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Example Question #2 : Using Expressions With Complex Numbers

Simplify the following complex number expression:

$(\sqrt{-8})(\sqrt{-6})$

Possible Answers:

$4\sqrt{3}$

$-4\sqrt{3}$

$-3\sqrt{2}$

$-4\sqrt{2}$

$4\sqrt{2}$

Correct answer:

$-4\sqrt{3}$

Explanation:

Begin by factoring the radical expression using imaginary numbers:

$(\sqrt{-8})(\sqrt{-6})$

$(2i\sqrt{2})(i\sqrt{6})$

Now, multiply the factors:

$(2i\sqrt{2})(i\sqrt{6}) = 2i^2\sqrt{12}$

Simplify. Remember that i^2 is equivalent to

$4i^2\sqrt{3}$

$-4\sqrt{3}$

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Example Question #4 : Using Expressions With Complex Numbers

Simplify the following complex number expression:

(3i^3)(-4i)^2

Possible Answers:

24i

36i

40i

48i

30i

Correct answer:

48i

Explanation:

Begin by completing the square:

(3i^3)(-4i)^2

(3i^3)(16i^2)

Now, multiply the factors. Remember that i^5 is equivalent to

(3i^3)(16i^2)=48i^5

48i

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High School Math : High School Math

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #6 : Imaginary Numbers

Example Question #1221 : High School Math

Example Question #5 : Imaginary Numbers

Example Question #2021 : Mathematical Relationships And Basic Graphs

Example Question #1 : Imaginary Numbers

Example Question #1222 : High School Math

Example Question #2 : Using Expressions With Complex Numbers

Example Question #1 : Using Expressions With Complex Numbers

Example Question #2 : Using Expressions With Complex Numbers

Example Question #4 : Using Expressions With Complex Numbers

All High School Math Resources

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