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ACT Math : Exponents

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Example Question #1 : How To Divide Complex Numbers

Complex numbers take the form a+bi , where is the real term in the complex number and is the nonreal (imaginary) term in the complex number.

Simplify: $\frac{6+12i}{4}$

Possible Answers:

$\frac{3}{2} + \frac{12i}{4i}$

6 + 3i

$\textup{None of these answers are correct}$

$\frac{3}{2} + 12i$

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

Correct answer:

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

Explanation:

This problem can be solved very similarly to a binomial such as a+bx . In this case, both the real and nonreal terms in the complex number are eligible to be divided by the real divisor.

$\frac{6+12i}{4} = (\frac{6}{4} + \frac{12i}{4})$

$\frac{12i}{4} = \frac{12}{4} \cdot i = 3 \cdot i = 3i$ , so

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

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Example Question #7 : How To Divide Complex Numbers

Complex numbers take the form a+bi , where is the real term in the complex number and is the nonreal (imaginary) term in the complex number.

Simplify by using conjugates: $\frac{4+2i}{-3-2i}$

Possible Answers:

$\frac{-16-2i}{13}$

$\frac{-16+2i}{-13}$

$\frac{-8-2i}{5}$

$\frac{-8-2i}{13}$

$\frac{-16-2i}{5}$

Correct answer:

$\frac{-16-2i}{13}$

Explanation:

Solving this problem using a conjugate is just like conjugating a binomial to simplify a denominator.

$\frac{(4+2i)(-3+2i)}{(-3-2i)(-3+2i)}$ Multiply both terms by the denominator's conjugate.

$\frac{(4+2i)(-3+2i)}{(-3-2i)(-3+2i)} = \frac{(4+2i)(-3+2i)}{13}$ Simplify. Note .

Combine and simplify.

Simplify the numerator.

$\frac{-16-2i}{13}$ The prime denominator prevents further simplifying.

Thus, $\frac{4+2i}{-3-2i} = \frac{-16-2i}{13}$ .

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Example Question #8 : How To Divide Complex Numbers

Simplify:

${}\frac{5+7i}{2+i}$

Possible Answers:

$\frac{5}{7}+2i{}$

$\frac{5}{2}+\frac{7}{2}i{}$

$\frac{17}{5}+\frac{9}{5}i{}$

1+3i{}

Correct answer:

$\frac{17}{5}+\frac{9}{5}i{}$

Explanation:

This problem can be solved in a way similar to other kinds of division problems (with binomials, for example). We need to get the imaginary number out of the denominator, so we will multiply the denominator by its conjugate and multiply the top by it as well to preserve the number's value.

$\frac{5+7i}{2+i} \cdot \frac{2-i}{2-i}=\frac{10-5i+14i-7i^2}{4-i^2}{}$

Then, recall i^2 = -1 by definition, so we can simplify this further:

$\frac{10+7+14i-5i}{4+1} = \frac{17}{5}+\frac{9}{5}i{}$

This is as far as we can simplify, so it is our final answer.

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Example Question #1 : How To Multiply Complex Numbers

The solution of $\sqrt{x-3}>2$ is the set of all real numbers such that:

Possible Answers:

x>5

x>3

x<7

x>1

x>7

Correct answer:

x>7

Explanation:

Square both sides of the equation: $x-3=2^{2}$

Then Solve for x: x=7

Therefore, x>7

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Example Question #2 : How To Multiply Complex Numbers

What is the product of 7i-3 and 6+2i

Possible Answers:

-18+36i

-4+36i

-32+36i

-32+48i

-18+14i

Correct answer:

-32+36i

Explanation:

Multiplying complex numbers is like multiplying binomials, you have to use foil. The only difference is, when you multiply the two terms that have in the them you can simplify the i^2 to negative 1. Foil is first, outside, inside, last

First

$7i\cdot6=42i$

Outside:

$7i\cdot2i=14i^2=-14$

Inside

$-3\cdot6=-18$

Last

$-3\cdot2i=-6i$

Add them all up and you get -32+36i

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Example Question #3 : How To Multiply Complex Numbers

Simplify the following:

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

Possible Answers:

Correct answer:

Explanation:

Begin this problem by doing a basic FOIL, treating just like any other variable. Thus, you know:

(9i + 12)(3i-4)=27i^2-36i+36i-48=27i^2-48

Recall that since $i=\sqrt{-1}$ , i^2=-1 . Therefore, you can simplify further:

27i^2-48 = 27(-1)-48=-75

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Example Question #4 : How To Multiply Complex Numbers

Complex numbers take the form a+bi , where is the real term in the complex number and is the nonreal (imaginary) term in the complex number.

Distribute: 2(4+7i)

Possible Answers:

8+7i

$\textup{None of these are correct}$

8+14i

6+7i

4+14i

Correct answer:

8+14i

Explanation:

This equation can be solved very similarly to a binomial like a+bx . Distribution takes place into both the real and nonreal terms inside the complex number, where applicable.

2(4+7i)

$(2\cdot4) + (2\cdot 7i)$

8 + 14i

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Example Question #5 : How To Multiply Complex Numbers

Complex numbers take the form a+bi , where is the real term in the complex number and is the nonreal (imaginary) term in the complex number.

Distribute and solve: 5(3+i) + 2(6+2i)

Possible Answers:

27 +3i

9 + 3i

27 + 9i

9 + 9i

$\textup{None of these answers are correct}$

Correct answer:

27 + 9i

Explanation:

This problem can be solved very similarly to a binomial like a+bx .

5(3+i) + 2(6+2i)

$((5\cdot3)+ (5 \cdot i)) + ((2 \cdot 6) + (2\cdot 2i))$

(15 + 5i) + (12 + 4i)

(15 + 12) + (5i + 4i)

27 + 9i

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Example Question #1 : How To Multiply Complex Numbers

Complex numbers take the form a+bi , where is the real term in the complex number and is the nonreal (imaginary) term in the complex number.

Which of the following is equivalent to i^3 ?

Possible Answers:

$i^{82}$

$i^{1000}$

$i^{30}$

$i^{23}$

Correct answer:

$i^{23}$

Explanation:

When dealing with complex numbers, remember that $i = \sqrt{-1}$ .

If we square , we thus get $i^2 = (\sqrt{-1})^2 = -1$ .

Yet another exponent gives us $i^3 = i^2 \cdot i = -1 \cdot i = -i$ OR $-\sqrt{-1}$ .

But when we hit i^4 , we discover that $i^4 = i^2 \cdot i^2 = -1 \cdot -1 = 1$

Thus, we have a repeating pattern with powers of , with every 4 exponents repeating the pattern. This means any power of evenly divisible by 4 will equal 1, any power of divisible by 4 with a remainder of 1 will equal , and so on.

Thus, $i^{23}$

$\frac{23}{4} = 5 \frac{3}{4}$

Since the remainder is 3, we know that $i^{23} = i^3 = -i$ .

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Example Question #7 : How To Multiply Complex Numbers

Simplify the following:

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

Possible Answers:

Correct answer:

Explanation:

Begin by treating this just like any normal case of FOIL. Notice that this is really the form of a difference of squares. Therefore, the distribution is very simple. Thus:

(4-3i)(4+3i)=16-9i^2

Now, recall that $i=\sqrt{-1}$ . Therefore, i^2 is . Based on this, we can simplify further:

16-9(-1)=16+9=25

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ACT Math : Exponents

Study concepts, example questions & explanations for ACT Math

All ACT Math Resources

Example Questions

Example Question #1 : How To Divide Complex Numbers

Example Question #7 : How To Divide Complex Numbers

Example Question #8 : How To Divide Complex Numbers

Example Question #1 : How To Multiply Complex Numbers

Example Question #2 : How To Multiply Complex Numbers

Example Question #3 : How To Multiply Complex Numbers

Example Question #4 : How To Multiply Complex Numbers

Example Question #5 : How To Multiply Complex Numbers

Example Question #1 : How To Multiply Complex Numbers

Example Question #7 : How To Multiply Complex Numbers

All ACT Math Resources

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