which expression is equivalent to $i^{233}$?\n1\n-1\n$i$\n-$i$

Explanation:

Step1: Recall the powers of $i$

The powers of the imaginary unit $i$ have a cyclic pattern: $i^1 = i$, $i^2=- 1$, $i^3 = i^2\times i=-i$, $i^4=(i^2)^2 = 1$. The cycle repeats every 4 powers.

Step2: Divide the exponent by 4

Divide 233 by 4. We use the division algorithm $233 = 4\times58+1$. So $i^{233}=i^{4\times58 + 1}$.

Step3: Apply the exponent - rule

Using the rule $a^{m + n}=a^m\times a^n$ and $(a^m)^n=a^{mn}$, we have $i^{4\times58+1}=(i^4)^{58}\times i^1$.

Step4: Evaluate the expression

Since $i^4 = 1$, then $(i^4)^{58}=1^{58}=1$. And $1\times i^1 = i$.

Answer:

C. $i$