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ACT Math : Complex Numbers

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Example Questions

Example Question #1 : How To Multiply Complex Numbers

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

Possible Answers:

-32+36i

-18+36i

-18+14i

-4+36i

-32+48i

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 #3 : 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+14i

$\textup{None of these are correct}$

8+7i

4+14i

6+7i

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

Possible Answers:

9 + 3i

27 + 9i

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

9 + 9i

27 +3i

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 #22 : 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 #3 : 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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Example Question #8 : How To Multiply Complex Numbers

Which of the following is equal to -16i^6 ?

Possible Answers:

16i

Correct answer:

Explanation:

Remember that since $i = \sqrt{-1}{}$ , you know that i^2 is . Therefore, i^4 is -1*-1 or . This makes our question very easy.

i^6 is the same as i^4*i^2 or 1 * -1

Thus, we know that -16i^6 is the same as -16 * -1 or .

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Example Question #41 : Squaring / Square Roots / Radicals

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 the following expression, leaving no complex numbers in the denominator.

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

Possible Answers:

$\frac{8+ 3i}{32}$

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

$\frac{20-4i}{15}$

$\frac{1+5i}{8}$

$\frac{-2+4i}{8}$

Correct answer:

$\frac{1+5i}{8}$

Explanation:

Solving this problem requires eliminating the nonreal term of the denominator. Our best bet for this is to cancel the nonreal term out by using the conjugate of the denominator.

Remember that for all binomials a+b , there exists a conjugate a-b such that $(a+b)\cdot(a-b) = (a^2-b^2)$ .

This can also be applied to complex conjugates, which will eliminate the nonreal portion entirely (since i^2 =-1 )!

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

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

$\frac{(3+2i)(4+4i)}{(16-16i^2)} = \frac{(3+2i)(4+4i)}{32}$ Simplify. Note .

(3+2i)(4+4i) = (12 + 12i + 8i + 8i^2) FOIL the numerator.

Combine and simplify.

$\frac{4+20i}{32} = \frac{1+5i}{8}$ Simplify the fraction.

Thus, $\frac{3+2i}{4-4i} = \frac{1+5i}{8}$ .

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ACT Math : Complex Numbers

Study concepts, example questions & explanations for ACT Math

All ACT Math Resources

Example Questions

Example Question #1 : How To Multiply Complex Numbers

Example Question #3 : How To Multiply Complex Numbers

Example Question #3 : How To Multiply Complex Numbers

Example Question #4 : How To Multiply Complex Numbers

Example Question #22 : Complex Numbers

Example Question #3 : How To Multiply Complex Numbers

Example Question #8 : How To Multiply Complex Numbers

Example Question #41 : Squaring / Square Roots / Radicals

All ACT Math Resources

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