Answer :
To determine the value of [tex]\(i^{233}\)[/tex], let's analyze the behavior of the imaginary unit [tex]\(i\)[/tex]. The imaginary unit [tex]\(i\)[/tex] has the property that [tex]\(i^2 = -1\)[/tex]. Additionally, the powers of [tex]\(i\)[/tex] follow a repeating pattern every four steps:
[tex]\[ \begin{aligned} i^1 &= i, \\ i^2 &= -1, \\ i^3 &= -i, \\ i^4 &= 1. \end{aligned} \][/tex]
This cycle repeats every four powers. Therefore, any power of [tex]\(i\)[/tex] can be reduced by finding the remainder when the exponent is divided by 4.
Let's perform this operation with the exponent 233:
[tex]\[ 233 \div 4 = 58 \text{ remainder } 1 \][/tex]
So, we can write 233 as:
[tex]\[ 233 = 4 \times 58 + 1 \][/tex]
This tells us that:
[tex]\[ i^{233} \equiv i^1 \pmod{4} \][/tex]
From our known pattern, [tex]\(i^1 = i\)[/tex].
Thus, [tex]\(i^{233} = i\)[/tex].
So, the expression equivalent to [tex]\(i^{233}\)[/tex] is:
[tex]\[ \boxed{i} \][/tex]
[tex]\[ \begin{aligned} i^1 &= i, \\ i^2 &= -1, \\ i^3 &= -i, \\ i^4 &= 1. \end{aligned} \][/tex]
This cycle repeats every four powers. Therefore, any power of [tex]\(i\)[/tex] can be reduced by finding the remainder when the exponent is divided by 4.
Let's perform this operation with the exponent 233:
[tex]\[ 233 \div 4 = 58 \text{ remainder } 1 \][/tex]
So, we can write 233 as:
[tex]\[ 233 = 4 \times 58 + 1 \][/tex]
This tells us that:
[tex]\[ i^{233} \equiv i^1 \pmod{4} \][/tex]
From our known pattern, [tex]\(i^1 = i\)[/tex].
Thus, [tex]\(i^{233} = i\)[/tex].
So, the expression equivalent to [tex]\(i^{233}\)[/tex] is:
[tex]\[ \boxed{i} \][/tex]
