5. $(-3i^7)^4 cdot i^{16}$
6. $(1 + 5i)(4 - 2i)$
7. $(7 - i)^2$
8. $\frac{8}{-12i}$
9. $\frac{3 - 6i}{4 - 3i}$
simplify, then name all sets to which the value belongs.
10. $i^{59}$
11. $(9 + 5i)(9 - 5i)$
for questions 11 and 12, use the complex numbers to write an example of each property.
12. associative property of multiplication
13. distributive property
14. name the additive identity of $(-10 + 4i)$.
15. name the multiplicative inverse of $7i$.
$-\frac{1}{7}i$
16. name all sets that are closed under subtraction.
$c, r, z, q$
© gina wilson (all things algebra®, llc), 2015-20

Question

Accepted Answer

### Question 5: $(-3i^7)^4 \cdot i^{16}$
# Explanation:
## Step 1: Simplify $i^7$
Recall that $i^1 = i$, $i^2 = -1$, $i^3 = -i$, $i^4 = 1$, and the powers of $i$ repeat every 4. So, $i^7 = i^{4 + 3} = i^4 \cdot i^3 = 1 \cdot (-i) = -i$.
## Step 2: Simplify $(-3i^7)^4$
Substitute $i^7 = -i$ into the expression: $(-3(-i))^4 = (3i)^4$. Using the power of a product rule $(ab)^n = a^n b^n$, we get $3^4 \cdot i^4 = 81 \cdot 1 = 81$ (since $i^4 = 1$).
## Step 3: Simplify $i^{16}$
Since $16$ is a multiple of $4$, $i^{16} = (i^4)^4 = 1^4 = 1$.
## Step 4: Multiply the two results
Multiply $81$ (from $(-3i^7)^4$) and $1$ (from $i^{16}$): $81 \cdot 1 = 81$.
# Answer:
$81$

### Question 6: $(1 + 5i)(4 - 2i)$
# Explanation:
## Step 1: Use the distributive property (FOIL method)
Multiply each term in the first complex number by each term in the second complex number:
$$
\begin{align*}
(1)(4) + (1)(-2i) + (5i)(4) + (5i)(-2i) &= 4 - 2i + 20i - 10i^2
\end{align*}
$$
## Step 2: Simplify $i^2$ and combine like terms
Recall that $i^2 = -1$, so substitute that in:
$$
\begin{align*}
4 - 2i + 20i - 10(-1) &= 4 - 2i + 20i + 10\
&= (4 + 10) + (-2i + 20i)\
&= 14 + 18i
\end{align*}
$$
# Answer:
$14 + 18i$

### Question 7: $(7 - i)^2$
# Explanation:
## Step 1: Use the formula $(a - b)^2 = a^2 - 2ab + b^2$
Here, $a = 7$ and $b = i$. So,
$$
\begin{align*}
(7 - i)^2 &= 7^2 - 2(7)(i) + i^2\
&= 49 - 14i + i^2
\end{align*}
$$
## Step 2: Simplify $i^2$
Since $i^2 = -1$, substitute that in:
$$
\begin{align*}
49 - 14i + (-1) &= (49 - 1) - 14i\
&= 48 - 14i
\end{align*}
$$
# Answer:
$48 - 14i$

### Question 8: $\frac{8}{-12i}$
# Explanation:
## Step 1: Multiply numerator and denominator by $i$ to rationalize the denominator
$$
\frac{8}{-12i} \cdot \frac{i}{i} = \frac{8i}{-12i^2}
$$
## Step 2: Simplify $i^2$
Since $i^2 = -1$, substitute that in:
$$
\frac{8i}{-12(-1)} = \frac{8i}{12}
$$
## Step 3: Simplify the fraction
Simplify $\frac{8i}{12}$ by dividing numerator and denominator by 4: $\frac{2i}{3}$.
# Answer:
$\frac{2i}{3}$

### Question 9: $\frac{3 - 6i}{4 - 3i}$
# Explanation:
## Step 1: Multiply numerator and denominator by the conjugate of the denominator
The conjugate of $4 - 3i$ is $4 + 3i$. So,
$$
\frac{3 - 6i}{4 - 3i} \cdot \frac{4 + 3i}{4 + 3i} = \frac{(3 - 6i)(4 + 3i)}{(4 - 3i)(4 + 3i)}
$$
## Step 2: Simplify the numerator
Use the distributive property (FOIL method) on the numerator:
$$
\begin{align*}
(3)(4) + (3)(3i) - 6i(4) - 6i(3i) &= 12 + 9i - 24i - 18i^2\
&= 12 - 15i - 18(-1)\
&= 12 - 15i + 18\
&= 30 - 15i
\end{align*}
$$
## Step 3: Simplify the denominator
Use the difference of squares formula $(a - b)(a + b) = a^2 - b^2$:
$$
4^2 - (3i)^2 = 16 - 9i^2 = 16 - 9(-1) = 16 + 9 = 25
$$
## Step 4: Simplify the fraction
Divide the numerator by the denominator: $\frac{30 - 15i}{25} = \frac{30}{25} - \frac{15i}{25} = \frac{6}{5} - \frac{3i}{5}$.
# Answer:
$\frac{6}{5} - \frac{3i}{5}$

### Question 10: $i^{59}$
# Explanation:
## Step 1: Find the remainder when 59 is divided by 4
Divide 59 by 4: $59 = 4 \times 14 + 3$. So, the remainder is 3.
## Step 2: Simplify $i^{59}$
Since the remainder is 3, $i^{59} = i^{4 \times 14 + 3} = (i^4)^{14} \cdot i^3$. We know that $i^4 = 1$ and $i^3 = -i$, so:
$$
(i^4)^{14} \cdot i^3 = 1^{14} \cdot (-i) = -i
$$
# Answer:
$-i$

### Question 11: $(9 + 5i)(9 - 5i)$
# Explanation:
## Step 1: Use the difference of squares formula
The formula $(a + b)(a - b) = a^2 - b^2$ applies here, where $a = 9$ and $b = 5i$. So,
$$
(9 + 5i)(9 - 5i) = 9^2 - (5i)^2
$$
## Step 2: Simplify the terms
Calculate $9^2 = 81$ and $(5i)^2 = 25i^2$. Since $i^2 = -1$, substitute that in:
$$
81 - 25(-1) = 81 + 25 = 106
$$
# Answer:
$106$

### Question 12: Associative Property of Multiplication (using complex numbers)
The associative property of multiplication states that for any complex numbers $a$, $b$, and $c$, $(a \cdot b) \cdot c = a \cdot (b \cdot c)$.

Let $a = 1 + i$, $b = 2 - i$, and $c = 3 + 2i$.

First, calculate $(a \cdot b) \cdot c$:
- $a \cdot b = (1 + i)(2 - i) = 2 - i + 2i - i^2 = 2 + i - (-1) = 3 + i$
- Then, $(a \cdot b) \cdot c = (3 + i)(3 + 2i) = 9 + 6i + 3i + 2i^2 = 9 + 9i + 2(-1) = 7 + 9i$

Next, calculate $a \cdot (b \cdot c)$:
- $b \cdot c = (2 - i)(3 + 2i) = 6 + 4i - 3i - 2i^2 = 6 + i - 2(-1) = 8 + i$
- Then, $a \cdot (b \cdot c) = (1 + i)(8 + i) = 8 + i + 8i + i^2 = 8 + 9i + (-1) = 7 + 9i$

Since $(a \cdot b) \cdot c = a \cdot (b \cdot c)$, the associative property holds.

# Answer:
An example is $((1 + i)(2 - i))(3 + 2i) = (1 + i)((2 - i)(3 + 2i))$, both equal to $7 + 9i$.

### Question 13: Distributive Property (using complex numbers)
The distributive property states that for any complex numbers $a$, $b$, and $c$, $a \cdot (b + c) = a \cdot b + a \cdot c$.

Let $a = 2 + i$, $b = 3 - 2i$, and $c = 4 + 3i$.

Calculate $a \cdot (b + c)$:
- First, find $b + c$: $(3 - 2i) + (4 + 3i) = 7 + i$
- Then, $a \cdot (b + c) = (2 + i)(7 + i) = 14 + 2i + 7i + i^2 = 14 + 9i + (-1) = 13 + 9i$

Calculate $a \cdot b + a \cdot c$:
- $a \cdot b = (2 + i)(3 - 2i) = 6 - 4i + 3i - 2i^2 = 6 - i - 2(-1) = 8 - i$
- $a \cdot c = (2 + i)(4 + 3i) = 8 + 6i + 4i + 3i^2 = 8 + 10i + 3(-1) = 5 + 10i$
- Then, $a \cdot b + a \cdot c = (8 - i) + (5 + 10i) = 13 + 9i$

Since $a \cdot (b + c) = a \cdot b + a \cdot c$, the distributive property holds.

# Answer:
An example is $(2 + i)((3 - 2i) + (4 + 3i)) = (2 + i)(3 - 2i) + (2 + i)(4 + 3i)$, both equal to $13 + 9i$.

### Question 14: Additive identity of $(-10 + 4i)$
The additive identity of any complex number $a + bi$ is $0 + 0i$ (or simply $0$) because adding $0$ to a complex number does not change it. So, $(-10 + 4i) + 0 = -10 + 4i$.

# Answer:
$0$ (or $0 + 0i$)

### Question 15: Multiplicative inverse of $7i$
The multiplicative inverse of a complex number $z$ is a complex number $w$ such that $z \cdot w = 1$. Let $z = 7i$ and $w = a + bi$. Then:

$$
7i \cdot (a + bi) = 1
$$

## Step 1: Expand the left side
$$
7i \cdot a + 7i \cdot bi = 7ai + 7bi^2
$$

## Step 2: Simplify $i^2$
Since $i^2 = -1$, substitute that in:
$$
7ai + 7b(-1) = -7b + 7ai
$$

## Step 3: Set equal to 1 (which is $1 + 0i$)
We have the system of equations:
- Real part: $-7b = 1$
- Imaginary part: $7a = 0$

From the imaginary part equation, $a = 0$. From the real part equation, $b = -\frac{1}{7}$. So, the multiplicative inverse is $0 - \frac{1}{7}i = -\frac{1}{7}i$.

# Answer:
$-\frac{1}{7}i$

### Question 16: Name all sets that are closed under subtraction
A set is closed under subtraction if the difference of any two elements in the set is also in the set.

- **Complex numbers ($\mathbb{C}$):** For any two complex numbers $a + bi$ and $c + di$, their difference is $(a - c) + (b - d)i$, which is also a complex number.
- **Real numbers ($\mathbb{R}$):** For any two real numbers $a$ and $b$, their difference $a - b$ is also a real number.
- **Rational numbers ($\mathbb{Q}$):** For any two rational numbers $\frac{a}{b}$ and $\frac{c}{d}$ (where $b, d
eq 0$), their difference $\frac{a}{b} - \frac{c}{d} = \frac{ad - bc}{bd}$ is also a rational number (since $ad - bc$ and $bd$ are integers and $bd
eq 0$).
- **Integers ($\mathbb{Z}$):** For any two integers $a$ and $b$, their difference $a - b$ is also an integer.

# Answer:
The sets closed under subtraction are complex numbers ($\mathbb{C}$), real numbers ($\mathbb{R}$), rational numbers ($\mathbb{Q}$), and integers ($\mathbb{Z}$).

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QUESTION IMAGE

Question

Explanation:

Question 5: $(-3i^7)^4 \cdot i^{16}$

Step 1: Simplify $i^7$

Step 2: Simplify $(-3i^7)^4$

Step 3: Simplify $i^{16}$

Step 4: Multiply the two results

Step 1: Use the distributive property (FOIL method)

Step 2: Simplify $i^2$ and combine like terms

Step 1: Use the formula $(a - b)^2 = a^2 - 2ab + b^2$

Step 2: Simplify $i^2$

Answer:

Question 6: $(1 + 5i)(4 - 2i)$