Question

compute $i^{-118}$.

Explanation:

Step1: Rewrite negative exponent

$i^{-118} = \frac{1}{i^{118}}$

Step2: Simplify $i^{118}$ using cycle

Recall $i^4=1$, divide 118 by 4: $118 = 4\times29 + 2$. So $i^{118}=(i^4)^{29}\times i^2 = 1^{29}\times i^2 = i^2$

Step3: Substitute $i^2=-1$

$\frac{1}{i^2} = \frac{1}{-1} = -1$

Answer:

$-1$