Question

simplify.
i^{99}

• i
• 1

1
i

Explanation:

Step1: Recall the powers of $i$ cycle

The powers of $i$ 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 99 by 4: $99\div4 = 24$ with a remainder. Using the division formula $99 = 4\times24+3$.

Step3: Rewrite $i^{99}$

$i^{99}=i^{4\times24 + 3}$. According to the exponent - rule $a^{m + n}=a^m\times a^n$ and $(a^m)^n=a^{mn}$, we have $i^{4\times24+3}=(i^4)^{24}\times i^3$.

Step4: Simplify the expression

Since $i^4 = 1$, then $(i^4)^{24}=1^{24}=1$. And $i^3=-i$. So $(i^4)^{24}\times i^3=1\times(-i)=-i$.

Answer:

$-i$