3. Find the value of the followings: $$ \begin{array}{llll} ext { a. } i^{16} & ext { b. } i^{25} & ext { c. } i^{47} & ext { d. } i^{522} \ ext { e. } i^{4 n+1}, n \in \mathbb{Z}^{+} & ext {f. } i^{9}-3 i^{17} & ext { g. } i^{7}+\frac{1}{i^{4}} & ext { h. } i^{-22}\end{array} $$

Question

3. Find the value of the followings: $$ \begin{array}{llll}	ext { a. } i^{16} & 	ext { b. } i^{25} & 	ext { c. } i^{47} & 	ext { d. } i^{522} \ 	ext { e. } i^{4 n+1}, n \in \mathbb{Z}^{+} & 	ext {f. } i^{9}-3 i^{17} & 	ext { g. } i^{7}+\frac{1}{i^{4}} & 	ext { h. } i^{-22}\end{array} $$

UpStudy AI · Accepted Answer

To solve the problems involving powers of $$ i $$, we first recall that $$ i $$ is the imaginary unit defined as $$ i = \sqrt{-1} $$. The powers of $$ i $$ cycle every four terms:

$$
\begin{align*}
i^1 & = i, \
i^2 & = -1, \
i^3 & = -i, \
i^4 & = 1.
\end{align*}
$$

From this, we can deduce that:

$$
i^{n} = i^{n \mod 4}.
$$

Now, let's calculate each part:

### a. $$ i^{16} $$
$$
16 \mod 4 = 0 \quad \Rightarrow \quad i^{16} = i^0 = 1.
$$

### b. $$ i^{25} $$
$$
25 \mod 4 = 1 \quad \Rightarrow \quad i^{25} = i^1 = i.
$$

### c. $$ i^{47} $$
$$
47 \mod 4 = 3 \quad \Rightarrow \quad i^{47} = i^3 = -i.
$$

### d. $$ i^{522} $$
$$
522 \mod 4 = 2 \quad \Rightarrow \quad i^{522} = i^2 = -1.
$$

### e. $$ i^{4n+1}, n \in \mathbb{Z}^{+} $$
For any positive integer $$ n $$:
$$
4n + 1 \mod 4 = 1 \quad \Rightarrow \quad i^{4n+1} = i^1 = i.
$$

### f. $$ i^{9} - 3i^{17} $$
Calculating each term:
$$
9 \mod 4 = 1 \quad \Rightarrow \quad i^{9} = i^1 = i,
$$
$$
17 \mod 4 = 1 \quad \Rightarrow \quad i^{17} = i^1 = i.
$$
Thus,
$$
i^{9} - 3i^{17} = i - 3i = -2i.
$$

### g. $$ i^{7} + \frac{1}{i^{4}} $$
Calculating each term:
$$
7 \mod 4 = 3 \quad \Rightarrow \quad i^{7} = i^3 = -i,
$$
$$
i^{4} = 1 \quad \Rightarrow \quad \frac{1}{i^{4}} = \frac{1}{1} = 1.
$$
Thus,
$$
i^{7} + \frac{1}{i^{4}} = -i + 1 = 1 - i.
$$

### h. $$ i^{-22} $$
Calculating:
$$
-22 \mod 4 = 2 \quad \Rightarrow \quad i^{-22} = \frac{1}{i^{22}}.
$$
Now, calculate $$ i^{22} $$:
$$
22 \mod 4 = 2 \quad \Rightarrow \quad i^{22} = i^2 = -1.
$$
Thus,
$$
i^{-22} = \frac{1}{-1} = -1.
$$

### Summary of Results:
$$
\begin{array}{ll}
\text{a. } & 1 \
\text{b. } & i \
\text{c. } & -i \
\text{d. } & -1 \
\text{e. } & i \
\text{f. } & -2i \
\text{g. } & 1 - i \
\text{h. } & -1 \
\end{array}
$$
a. 1, b. i, c. -i, d. -1, e. i, f. -2i, g. 1 - i, h. -1

Find the value of the followings:

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