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Simplify the power of [tex]\( i \)[/tex].

[tex]\( i^{15} = \square \)[/tex]


Sagot :

To simplify powers of [tex]\(i\)[/tex], we need to understand the cyclical nature of imaginary unit [tex]\(i\)[/tex]. The powers of [tex]\(i\)[/tex] repeat every four steps:

[tex]\[ i^1 = i \][/tex]
[tex]\[ i^2 = -1 \][/tex]
[tex]\[ i^3 = -i \][/tex]
[tex]\[ i^4 = 1 \][/tex]

After [tex]\(i^4\)[/tex], the powers repeat the same pattern:
[tex]\[ i^5 = i \][/tex]
[tex]\[ i^6 = -1 \][/tex]
[tex]\[ i^7 = -i \][/tex]
[tex]\[ i^8 = 1 \][/tex]
...and so on.

To simplify higher powers of [tex]\(i\)[/tex], we can use the property of modular arithmetic. Specifically, for any integer [tex]\(n\)[/tex], [tex]\(i^n\)[/tex] can be simplified by finding the remainder when [tex]\(n\)[/tex] is divided by 4. This remainder will tell us where we are in the cycle.

For [tex]\(i^{15}\)[/tex]:

1. Calculate the remainder when 15 is divided by 4:
[tex]\[ 15 \div 4 = 3 \quad \text{remainder} \quad 3 \][/tex]

This indicates that [tex]\(i^{15}\)[/tex] corresponds to [tex]\(i^3\)[/tex] in our cycle.

2. Refer to our established cycle:
[tex]\[ i^3 = -i \][/tex]

Therefore,
[tex]\[ i^{15} = -i \][/tex]
Thus:

[tex]\[ \boxed{-i} \][/tex]
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