Certainly! To simplify [tex]\( i^{31} \)[/tex], we need to understand the cyclical nature of the powers of the imaginary unit [tex]\( i \)[/tex].
The powers of [tex]\( i \)[/tex] follow a repeating cycle every four steps:
1. [tex]\( i^1 = i \)[/tex]
2. [tex]\( i^2 = -1 \)[/tex]
3. [tex]\( i^3 = -i \)[/tex]
4. [tex]\( i^4 = 1 \)[/tex]
After [tex]\( i^4 = 1 \)[/tex], the powers repeat this cycle:
- [tex]\( i^5 = i \)[/tex]
- [tex]\( i^6 = -1 \)[/tex]
- [tex]\( i^7 = -i \)[/tex]
- [tex]\( i^8 = 1 \)[/tex]
This pattern continues indefinitely. To determine [tex]\( i^{31} \)[/tex], we need to find where 31 falls within this cycle.
We do this by taking the exponent modulo 4:
[tex]\[ 31 \mod 4 \][/tex]
Calculating this, we find:
[tex]\[ 31 \div 4 = 7 \quad \text{remainder} \quad 3 \][/tex]
Hence,
[tex]\[ 31 \mod 4 = 3 \][/tex]
This tells us that [tex]\( i^{31} \)[/tex] is equivalent to [tex]\( i^3 \)[/tex] in the cycle.
From the cycle we know:
[tex]\[ i^3 = -i \][/tex]
Thus, the simplified form of [tex]\( i^{31} \)[/tex] is:
[tex]\[ -i \][/tex]
So the answer is:
[tex]\[ i^{31} = -i \][/tex]
