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ACT Math : How to multiply complex numbers

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Example Questions

Example Question #31 :Squaring / Square Roots / Radicals

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

Possible Answers:

x<7

x>1

x>3

x>5

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 #32 :Squaring / Square Roots / Radicals

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

Possible Answers:

-4+36i

-18+14i

-18+36i

-32+36i

-32+48i

Correct answer:

-32+36i

Explanation:

用复数就像乘法binomials, you have to use foil. The only difference is, when you multiply the two terms that havein 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, treatingjust 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 #21 :Complex Numbers

Complex numbers take the form a+bi , whereis the real term in the complex number andis the nonreal (imaginary) term in the complex number.

Distribute: 2(4+7i)

Possible Answers:

8+14i

4+14i

8+7i

$\textup{None of these are correct}$

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 #5 :How To Multiply Complex Numbers

Complex numbers take the form a+bi , whereis the real term in the complex number andis the nonreal (imaginary) term in the complex number.

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

Possible Answers:

9 + 3i

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

27 +3i

9 + 9i

27 + 9i

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 #6 :How To Multiply Complex Numbers

Complex numbers take the form a+bi , whereis the real term in the complex number andis the nonreal (imaginary) term in the complex number.

Which of the following is equivalent to i^3 ?

Possible Answers:

$i^{1000}$

$i^{82}$

$i^{23}$

$i^{30}$

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 ofevenly divisible by 4 will equal 1, any power ofdivisible 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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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 #2351 :Act Math

Complex numbers take the form a+bi , whereis the real term in the complex number andis 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:

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

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

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

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

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

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 theconjugateof 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 : How to multiply complex numbers

Study concepts, example questions & explanations for ACT Math

All ACT Math Resources

Example Questions

Example Question #31 :Squaring / Square Roots / Radicals

Example Question #32 :Squaring / Square Roots / Radicals

Example Question #3 :How To Multiply Complex Numbers

Example Question #21 :Complex Numbers

Example Question #5 :How To Multiply Complex Numbers

Example Question #6 :How To Multiply Complex Numbers

Example Question #7 :How To Multiply Complex Numbers

Example Question #8 :How To Multiply Complex Numbers

Example Question #2351 :Act Math

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