Imaginary Numbers

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Algebra II › Imaginary Numbers

Questions 1 - 10

1

Evaluate:

$| 9- i\sqrt{6}|$

$\sqrt{87}$

$5 \sqrt{3}$

$9+ \sqrt{6}$

$9- \sqrt{6}$

None of these

Explanation

refers to the absolute value of a complex number , which can be calculated by evaluating $\sqrt{a^{2}+ b^{2}}$ . Setting $a= 9, b = \sqrt{6}$ , the value of this expression is

$| 9- i\sqrt{6}| = \sqrt{9^{2}+ (\sqrt{6})^{2}} = \sqrt{81+ 6} = \sqrt{87}$

2

Explanation

Nothing can be simplified in either parentheses, so the first step is to distribute the negative sign to the second parentheses

Then, you combine similar terms remembering that terms with i cannot combine with those with no i

3

Consider the following definitions of imaginary numbers:

Then,

Explanation

4

Explanation

This problem requires you to use FOIL to multiply the binomials.

Multiply the first terms

$(5i\cdot i=5i^2)$ ,

then the outside terms

$(5i\cdot 3=15i)$ ,

next the inside terms

$(-2\cdot i=-2i)$ ,

and finally the last terms

$(-2\cdot 3=-6)$ .

Put those together to get

.

Recall that

.

Therefore, your answer is

.

5

Solve: $-2i^2-4i^{-4}+6i^6$

Explanation

Evaluate each term of the expression. Write out the values of the imaginary terms.

$i=\sqrt{-1}$

$i^2= i\cdot i = -1$

$i^4 = i^2\cdot i^2 = (-1)\cdot (-1) = 1$

$i^6 = i^4 \cdot i^2 = (1)\cdot (-1) = -1$

Replace the values of each.

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

Sum all the values.

$-2i^2-4i^{-4}+6i^6 = 2-4-6 = -8$

The answer is:

6

$What\ is\ the\ complex\ conjugate\ for\ 9-2i?$

$81-4i^{2}$

$81+4i^{2}$

Explanation

$The\ complex\ conjugate\ for\ a binomial\ with\ an\ imaginary\ number\ is$

$simply\ the\ original\ with\ the\ sign\ for\ the\ imaginary\ number\ changed.$

$If\ 9-2i\ were\ multiplied\ by\ 9+2i\ the\ i\ values\ would\ cancel\ producing\ a\ real\ number.$

7

Evaluate: $\frac{i^6+1}{i-1}$

$\frac{1}{2}i-\frac{1}{2}$

Explanation

The imaginary term is equivalent to $\sqrt{-1}$ .

This means that:

Substitute this term back into the numerator.

$\frac{i^6+1}{i-1}=\frac{-1+1}{i-1} = \frac{0}{i-1} =0$

There is no need to use extra steps such as multiplying by the conjugate of the denominator to simplify.

The answer is:

8

Simplify: $\frac{i^{60}-1}{i^3-1}$

$\textup{Undefined.}$

Explanation

Write the first few terms of the imaginary term.

$i=\sqrt{-1}$

$i^2 = i\cdot i =-1$

$i^3 = i^2\cdot i = -i$

$i^4 = i^2\cdot i^2=1$

Notice that these terms will be in a pattern for higher order imaginary terms.

Rewrite the numerator using the product of exponents.

$\frac{i^{60}-1}{i^3-1} =\frac{(i^4)^{15}-1}{i^3-1} = \frac{1-1}{-i-1} = 0$

The answer is:

9

Which of the following is equal to $-\sqrt{-1}$ ?

1

Explanation

Each of the following are true:

$i = \sqrt{-1}$

$i^2 = (\sqrt{-1})^2 = -1$

$i^3 = (\sqrt{-1})^3 = -\sqrt{-1}$

$i^4 = (\sqrt{-1})^4 = 1$

Therefore, the correct answer is .

10

Evaluate:

Explanation

Write the powers of the imaginary numbers.

$i=\sqrt{-1}$

$i^2= i\cdot i = -1$

$i^3 = i^2\cdot i = -i$

$i^4 = i^2\cdot i^2 = (-1)\cdot (-1) = 1$

Notice that this will repeat. We can rewrite higher powers if the imaginary term by product of powers.

The answer is:

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