1. The problem is to simplify the expression $\frac{1}{i^{199}}$ where $i$ is the imaginary unit with the property $i^2 = -1$. 2. Recall the powers of $i$ cycle every 4 steps: $$i^1 = i, \quad i^2 = -1, \quad i^3 = -i, \quad i^4 = 1, \quad \text{and then it repeats}.$$ 3. To simplify $i^{199}$, find the remainder when 199 is divided by 4: $$199 \div 4 = 49 \text{ remainder } 3.$$ So, $$i^{199} = i^{4 \times 49 + 3} = (i^4)^{49} \times i^3 = 1^{49} \times i^3 = i^3 = -i.$$ 4. Substitute back into the original expression: $$\frac{1}{i^{199}} = \frac{1}{-i} = -\frac{1}{i}.$$ 5. To write $-\frac{1}{i}$ in standard form, multiply numerator and denominator by $i$: $$-\frac{1}{i} \times \frac{i}{i} = -\frac{i}{i^2} = -\frac{i}{-1} = i.$$ **Final answer:** $$\frac{1}{i^{199}} = i.$$