Answer :
To determine the value of [tex]\( i^{22} \)[/tex], we need to recognize the cyclical pattern of the powers of the imaginary unit [tex]\( i \)[/tex], which is defined such that [tex]\( i^2 = -1 \)[/tex]. The powers of [tex]\( i \)[/tex] cycle every 4 exponents as follows:
1. [tex]\( i^1 = i \)[/tex]
2. [tex]\( i^2 = -1 \)[/tex]
3. [tex]\( i^3 = -i \)[/tex]
4. [tex]\( i^4 = 1 \)[/tex]
5. [tex]\( i^5 = i \)[/tex] (and the cycle repeats)
Given this cyclical pattern, we can find [tex]\( i^{22} \)[/tex] by determining the position within the cycle. This is done by finding the remainder when the exponent (22 in this case) is divided by the length of the cycle (which is 4).
First, let’s determine the remainder when 22 is divided by 4:
[tex]\[ 22 \div 4 = 5 \text{ R } 2 \][/tex]
This tells us that 22 divided by 4 equals 5 with a remainder of 2.
Hence, [tex]\( i^{22} \)[/tex] is equivalent to [tex]\( i^2 \)[/tex] because of the cyclical nature of the powers of [tex]\( i \)[/tex].
Looking at the cycle:
- [tex]\( i^1 = i \)[/tex]
- [tex]\( i^2 = -1 \)[/tex]
- [tex]\( i^3 = -i \)[/tex]
- [tex]\( i^4 = 1 \)[/tex]
We see that:
[tex]\[ i^2 = -1 \][/tex]
Therefore, [tex]\( i^{22} = -1 \)[/tex].
1. [tex]\( i^1 = i \)[/tex]
2. [tex]\( i^2 = -1 \)[/tex]
3. [tex]\( i^3 = -i \)[/tex]
4. [tex]\( i^4 = 1 \)[/tex]
5. [tex]\( i^5 = i \)[/tex] (and the cycle repeats)
Given this cyclical pattern, we can find [tex]\( i^{22} \)[/tex] by determining the position within the cycle. This is done by finding the remainder when the exponent (22 in this case) is divided by the length of the cycle (which is 4).
First, let’s determine the remainder when 22 is divided by 4:
[tex]\[ 22 \div 4 = 5 \text{ R } 2 \][/tex]
This tells us that 22 divided by 4 equals 5 with a remainder of 2.
Hence, [tex]\( i^{22} \)[/tex] is equivalent to [tex]\( i^2 \)[/tex] because of the cyclical nature of the powers of [tex]\( i \)[/tex].
Looking at the cycle:
- [tex]\( i^1 = i \)[/tex]
- [tex]\( i^2 = -1 \)[/tex]
- [tex]\( i^3 = -i \)[/tex]
- [tex]\( i^4 = 1 \)[/tex]
We see that:
[tex]\[ i^2 = -1 \][/tex]
Therefore, [tex]\( i^{22} = -1 \)[/tex].
