Question

simplify each of the following powers of i.
$i^{15}=-i$ (complete)
$i^{32}=1$ (complete)
$i^{99}=$ (do) options: i, -i, 1, -1

Explanation:

Step1: Recall the pattern of powers of $i$

$i^1 = i$, $i^2=- 1$, $i^3 = i^2\times i=-i$, $i^4=(i^2)^2 = 1$. The powers of $i$ repeat every 4 powers.

Step2: Divide the exponent by 4

For $i^{99}$, divide 99 by 4. $99\div4 = 24$ with a remainder. Using the division - formula $a = bq + r$, where $a = 99$, $b = 4$, $q = 24$, and $r=3$. So $i^{99}=i^{4\times24 + 3}$.

Step3: Apply the exponent - rule $(a^m)^n=a^{mn}$

By the 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: Substitute the values of $i^4$ and $i^3$

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:

$-i$