Question

compute $i^{95}$.

Explanation:

Step1: Record the pattern of powers of \(i\)

The powers of the imaginary unit \(i\) have a cyclic pattern:
\(i^1 = i\)
\(i^2 = - 1\)
\(i^3 = i^2\times i=-1\times i = -i\)
\(i^4=(i^2)^2=(-1)^2 = 1\)
And then the pattern repeats every 4 powers: \(i^{4n}=1\), \(i^{4n + 1}=i\), \(i^{4n+2}=-1\), \(i^{4n + 3}=-i\) for any integer \(n\).

Step2: Divide the exponent 95 by 4 to find the remainder

We perform the division \(95\div4\):
\(95 = 4\times23+3\)
Here, the quotient \(n = 23\) and the remainder is 3.

Step3: Determine the value of \(i^{95}\) based on the remainder

Since the remainder is 3, we use the pattern for \(i^{4n + 3}\). Substituting \(n = 23\) into \(i^{4n+3}\), we get:
\(i^{95}=i^{4\times23 + 3}=-i\)

Answer:

\(-i\)