Step-by-step explanation:
To find the value of \( \frac{1}{i^{99}} \), we need to simplify \( i^{99} \) first. The imaginary unit \( i \) is defined as \( \sqrt{-1} \), and it follows a cyclical pattern in its powers:
\[ i^1 = i \]
\[ i^2 = -1 \]
\[ i^3 = -i \]
\[ i^4 = 1 \]
\[ i^5 = i \]
This pattern repeats every four powers. To find \( i^{99} \), we can determine the remainder when 99 is divided by 4:
\[ 99 \mod 4 = 3 \]
Thus, \( i^{99} = i^3 \). From the pattern, we know that \( i^3 = -i \). Therefore,
\[ \frac{1}{i^{99}} = \frac{1}{-i} \]
To simplify \(\frac{1}{-i}\), we multiply the numerator and the denominator by \(i\):
\[ \frac{1}{-i} \times \frac{i}{i} = \frac{i}{-i^2} = \frac{i}{1} = i \]
Thus, the value of \(\frac{1}{i^{99}}\) is \(i\).