Complex Numbers - ACT Math

Card 0 of 252

Question

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

Answer

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

Then Solve for x:

Therefore,

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Question

What is the product of and

Answer

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 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

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Question

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

Distribute:

Answer

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

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

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Question

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

Distribute and solve:

Answer

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

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

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Question

Complex numbers take the form , 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 ?

Answer

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 , 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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Question

Complex numbers take the form , 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.

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

Answer

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 , there exists a conjugate 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 )!

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

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

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

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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Question

Which of the following is equal to ?

Answer

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

is the same as or

Thus, we know that is the same as or .

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Question

Simplify the following:

Answer

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:

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

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Question

Simplify the following:

Answer

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

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

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Question

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

Answer

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

Then Solve for x:

Therefore,

Compare your answer with the correct one above

Question

What is the product of and

Answer

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 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

Compare your answer with the correct one above

Question

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

Distribute:

Answer

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

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

Compare your answer with the correct one above

Question

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

Distribute and solve:

Answer

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

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

Compare your answer with the correct one above

Question

Complex numbers take the form , 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 ?

Answer

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 , 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 $i^{23} = i^3 = -i$ .

Compare your answer with the correct one above

Question

Complex numbers take the form , 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.

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

Answer

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 , there exists a conjugate 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 )!

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

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

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

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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Question

Which of the following is equal to ?

Answer

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

is the same as or

Thus, we know that is the same as or .

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Question

Simplify the following:

Answer

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:

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

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Question

Simplify the following:

Answer

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

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

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Question

Simplify:

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

Answer

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 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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Question

Simplify: $\frac{9+4i}{4i}$

Answer

Multiply both numberator and denominator by :

$\frac{9+4i}{4i}$

$= \frac{\left (9+4i \right )\cdot i}{4i \cdot i}$

$= \frac{9i + 4i^{2}}{4 i^{2}}$

$= \frac{9i + 4 (-1)}{4 (-1)}$

$= \frac{-4+9i }{-4 }$

$= \frac{-4 }{-4 }+ \frac{ 9i }{-4 }$

$= 1- \frac{ 9 }{4 } i$

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