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

Explanation:

Step1: Recall the powers of $i$

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

Step2: Divide the exponent by 4

Divide 233 by 4: $233\div4 = 58\cdots\cdots1$. Here, the quotient is 58 and the remainder is 1.

Step3: Rewrite $i^{233}$

We can rewrite $i^{233}$ as $i^{4\times58 + 1}$. Using the property of exponents $a^{m + n}=a^m\times a^n$ and $(a^m)^n=a^{mn}$, we have $i^{4\times58+1}=(i^4)^{58}\times i^1$. Since $i^4 = 1$, then $(i^4)^{58}=1^{58}=1$. So, $(i^4)^{58}\times i^1=1\times i = i$.

Answer:

$i$